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  • Order of operations
  • Evaluating expressions
  • Simplifying algebraic expressions
  • Multi-step equations
  • Work word problems
  • Distance-rate-time word problems
  • Mixture word problems
  • Absolute value equations
  • Multi-step inequalities
  • Compound inequalities
  • Absolute value inequalities
  • Discrete relations
  • Continuous relations
  • Evaluating and graphing functions
  • Review of linear equations
  • Graphing absolute value functions
  • Graphing linear inequalities
  • Direct and inverse variation
  • Systems of two linear inequalities
  • Systems of two equations
  • Systems of two equations, word problems
  • Points in three dimensions
  • Systems of three equations, elimination
  • Systems of three equations, substitution
  • Basic matrix operations
  • Matrix multiplication
  • All matrix operations combined
  • Matrix inverses
  • Geometric transformations with matrices
  • Operations with complex numbers
  • Properties of complex numbers
  • Rationalizing imaginary denominators
  • Properties of parabolas
  • Vertex form
  • Graphing quadratic inequalities
  • Factoring quadratic expressions
  • Solving quadratic equations w/ square roots
  • Solving quadratic equations by factoring
  • Completing the square
  • Solving equations by completing the square
  • Solving equations with the quadratic formula
  • The discriminant
  • Naming and simple operations
  • Factoring a sum/difference of cubes
  • Factoring by grouping
  • Factoring quadratic form
  • Factoring using all techniques
  • Factors and Zeros
  • The Remainder Theorem
  • Irrational and Imaginary Root Theorems
  • Descartes' Rule of Signs
  • More on factors, zeros, and dividing
  • The Rational Root Theorem
  • Polynomial equations
  • Basic shape of graphs of polynomials
  • Graphing polynomial functions
  • The Binomial Theorem
  • Evaluating functions
  • Function operations
  • Inverse functions
  • Simplifying radicals
  • Operations with radical expressions
  • Dividing radical expressions
  • Radicals and rational exponents
  • Simplifying rational exponents
  • Square root equations
  • Rational exponent equations
  • Graphing radicals
  • Graphing & properties of parabolas
  • Equations of parabolas
  • Graphing & properties of circles
  • Equations of circles
  • Graphing & properties of ellipses
  • Equations of ellipses
  • Graphing & properties of hyperbolas
  • Equations of hyperbolas
  • Classifying conic sections
  • Eccentricity
  • Systems of quadratic equations
  • Graphing simple rational functions
  • Graphing general rational functions
  • Simplifying rational expressions
  • Multiplying / dividing rational expressions
  • Adding / subtracting rational expressions
  • Complex fractions
  • Solving rational equations
  • The meaning of logarithms
  • Properties of logarithms
  • The change of base formula
  • Writing logs in terms of others
  • Logarithmic equations
  • Inverse functions and logarithms
  • Exponential equations not requiring logarithms
  • Exponential equations requiring logarithms
  • Graphing logarithms
  • Graphing exponential functions
  • Discrete exponential growth and decay word problems
  • Continuous exponential growth and decay word problems
  • General sequences
  • Arithmetic sequences
  • Geometric sequences
  • Comparing Arithmetic/Geometric Sequences
  • General series
  • Arithmetic series
  • Arithmetic/Geometric Means w/ Sequences
  • Finite geometric series
  • Infinite geometric series
  • Right triangle trig: Evaluating ratios
  • Right triangle trig: Missing sides/angles
  • Angles and angle measure
  • Co-terminal angles and reference angles
  • Arc length and sector area
  • Trig ratios of general angles
  • Exact trig ratios of important angles
  • The Law of Sines
  • The Law of Cosines
  • Graphing trig functions
  • Translating trig functions
  • Angle Sum/Difference Identities
  • Double-/Half-Angle Identities
  • Sample spaces and The Counting Principle
  • Independent and dependent events
  • Mutualy exclusive events
  • Permutations
  • Combinations
  • Permutations vs combinations
  • Probability using permutations and combinations

Graphing Quadratic Equations

Here is an example:

You can graph a Quadratic Equation using the Function Grapher , but to really understand what is going on, you can make the graph yourself. Read On!

The Simplest Quadratic

The simplest Quadratic Equation is:

And its graph is simple too:

This is the curve f(x) = x 2 It is a parabola .

Now let us see what happens when we introduce the "a" value:

f(x) = ax 2

  • Larger values of a squash the curve inwards
  • Smaller values of a expand it outwards
  • And negative values of a flip it upside down

The "General" Quadratic

Before graphing we rearrange the equation, from this:

f(x) = ax 2 + bx + c

f(x) = a(x-h) 2 + k

In other words, calculate h (= −b/2a), then find k by calculating the whole equation for x=h

The wonderful thing about this new form is that h and k show us the very lowest (or very highest) point, called the vertex :

And also the curve is symmetrical (mirror image) about the axis that passes through x=h , making it easy to graph

  • h shows us how far left (or right) the curve has been shifted from x=0
  • k shows us how far up (or down) the curve has been shifted from y=0

Lets see an example of how to do this:

Example: Plot f(x) = 2x 2 − 12x + 16

First, let's note down:

  • b = −12, and

Now, what do we know?

  • a is positive, so it is an "upwards" graph ("U" shaped)
  • a is 2, so it is a little "squashed" compared to the x 2 graph

Next, let's calculate h:

And next we can calculate k (using h=3):

So now we can plot the graph (with real understanding!):

We also know: the vertex is (3,−2), and the axis is x=3

From A Graph to The Equation

What if we have a graph, and want to find an equation?

Example: you have just plotted some interesting data, and it looks Quadratic:

Just knowing those two points we can come up with an equation.

Firstly, we know h and k (at the vertex):

(h, k) = (1, 1)

So let's put that into this form of the equation:

f(x) = a(x−1) 2 + 1

Then we calculate "a":

And here is the resulting Quadratic Equation:

f(x) = 0.5(x−1) 2 + 1

Note: This may not be the correct equation for the data, but it’s a good model and the best we can come up with.

  • 10.5 Graphing Quadratic Equations in Two Variables
  • Introduction
  • 1.1 Introduction to Whole Numbers
  • 1.2 Use the Language of Algebra
  • 1.3 Add and Subtract Integers
  • 1.4 Multiply and Divide Integers
  • 1.5 Visualize Fractions
  • 1.6 Add and Subtract Fractions
  • 1.7 Decimals
  • 1.8 The Real Numbers
  • 1.9 Properties of Real Numbers
  • 1.10 Systems of Measurement
  • Key Concepts
  • Review Exercises
  • Practice Test
  • 2.1 Solve Equations Using the Subtraction and Addition Properties of Equality
  • 2.2 Solve Equations using the Division and Multiplication Properties of Equality
  • 2.3 Solve Equations with Variables and Constants on Both Sides
  • 2.4 Use a General Strategy to Solve Linear Equations
  • 2.5 Solve Equations with Fractions or Decimals
  • 2.6 Solve a Formula for a Specific Variable
  • 2.7 Solve Linear Inequalities
  • 3.1 Use a Problem-Solving Strategy
  • 3.2 Solve Percent Applications
  • 3.3 Solve Mixture Applications
  • 3.4 Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem
  • 3.5 Solve Uniform Motion Applications
  • 3.6 Solve Applications with Linear Inequalities
  • 4.1 Use the Rectangular Coordinate System
  • 4.2 Graph Linear Equations in Two Variables
  • 4.3 Graph with Intercepts
  • 4.4 Understand Slope of a Line
  • 4.5 Use the Slope-Intercept Form of an Equation of a Line
  • 4.6 Find the Equation of a Line
  • 4.7 Graphs of Linear Inequalities
  • 5.1 Solve Systems of Equations by Graphing
  • 5.2 Solving Systems of Equations by Substitution
  • 5.3 Solve Systems of Equations by Elimination
  • 5.4 Solve Applications with Systems of Equations
  • 5.5 Solve Mixture Applications with Systems of Equations
  • 5.6 Graphing Systems of Linear Inequalities
  • 6.1 Add and Subtract Polynomials
  • 6.2 Use Multiplication Properties of Exponents
  • 6.3 Multiply Polynomials
  • 6.4 Special Products
  • 6.5 Divide Monomials
  • 6.6 Divide Polynomials
  • 6.7 Integer Exponents and Scientific Notation
  • 7.1 Greatest Common Factor and Factor by Grouping
  • 7.2 Factor Trinomials of the Form x2+bx+c
  • 7.3 Factor Trinomials of the Form ax2+bx+c
  • 7.4 Factor Special Products
  • 7.5 General Strategy for Factoring Polynomials
  • 7.6 Quadratic Equations
  • 8.1 Simplify Rational Expressions
  • 8.2 Multiply and Divide Rational Expressions
  • 8.3 Add and Subtract Rational Expressions with a Common Denominator
  • 8.4 Add and Subtract Rational Expressions with Unlike Denominators
  • 8.5 Simplify Complex Rational Expressions
  • 8.6 Solve Rational Equations
  • 8.7 Solve Proportion and Similar Figure Applications
  • 8.8 Solve Uniform Motion and Work Applications
  • 8.9 Use Direct and Inverse Variation
  • 9.1 Simplify and Use Square Roots
  • 9.2 Simplify Square Roots
  • 9.3 Add and Subtract Square Roots
  • 9.4 Multiply Square Roots
  • 9.5 Divide Square Roots
  • 9.6 Solve Equations with Square Roots
  • 9.7 Higher Roots
  • 9.8 Rational Exponents
  • 10.1 Solve Quadratic Equations Using the Square Root Property
  • 10.2 Solve Quadratic Equations by Completing the Square
  • 10.3 Solve Quadratic Equations Using the Quadratic Formula
  • 10.4 Solve Applications Modeled by Quadratic Equations

Learning Objectives

By the end of this section, you will be able to:

  • Recognize the graph of a quadratic equation in two variables
  • Find the axis of symmetry and vertex of a parabola
  • Find the intercepts of a parabola
  • Graph quadratic equations in two variables
  • Solve maximum and minimum applications

Be Prepared 10.13

Before you get started, take this readiness quiz.

Graph the equation y = 3 x − 5 y = 3 x − 5 by plotting points. If you missed this problem, review Example 4.11 .

Be Prepared 10.14

Evaluate 2 x 2 + 4 x − 1 2 x 2 + 4 x − 1 when x = −3 x = −3 . If you missed this problem, review Example 1.57 .

Be Prepared 10.15

Evaluate − b 2 a − b 2 a when a = 1 3 a = 1 3 and b = 5 6 b = 5 6 . If you missed this problem, review Example 1.89 .

Recognize the Graph of a Quadratic Equation in Two Variables

We have graphed equations of the form A x + B y = C A x + B y = C . We called equations like this linear equations because their graphs are straight lines.

Now, we will graph equations of the form y = a x 2 + b x + c y = a x 2 + b x + c . We call this kind of equation a quadratic equation in two variables .

Quadratic Equation in Two Variables

A quadratic equation in two variables , where a , b , and c a , b , and c are real numbers and a ≠ 0 a ≠ 0 , is an equation of the form

Just like we started graphing linear equations by plotting points, we will do the same for quadratic equations.

Let’s look first at graphing the quadratic equation y = x 2 y = x 2 . We will choose integer values of x x between −2 −2 and 2 and find their y y values. See Table 10.1 .

Notice when we let x = 1 x = 1 and x = −1 x = −1 , we got the same value for y y .

The same thing happened when we let x = 2 x = 2 and x = −2 x = −2 .

Now, we will plot the points to show the graph of y = x 2 y = x 2 . See Figure 10.2 .

The graph is not a line. This figure is called a parabola . Every quadratic equation has a graph that looks like this.

In Example 10.43 you will practice graphing a parabola by plotting a few points.

Example 10.43

Graph y = x 2 − 1 y = x 2 − 1 .

We will graph the equation by plotting points.

Try It 10.85

Graph y = − x 2 y = − x 2 .

Try It 10.86

Graph y = x 2 + 1 y = x 2 + 1 .

How do the equations y = x 2 y = x 2 and y = x 2 − 1 y = x 2 − 1 differ? What is the difference between their graphs? How are their graphs the same?

All parabolas of the form y = a x 2 + b x + c y = a x 2 + b x + c open upwards or downwards. See Figure 10.3 .

Notice that the only difference in the two equations is the negative sign before the x 2 x 2 in the equation of the second graph in Figure 10.3 . When the x 2 x 2 term is positive, the parabola opens upward, and when the x 2 x 2 term is negative, the parabola opens downward.

Parabola Orientation

For the quadratic equation y = a x 2 + b x + c y = a x 2 + b x + c , if:

Example 10.44

Determine whether each parabola opens upward or downward:

ⓐ y = −3 x 2 + 2 x − 4 y = −3 x 2 + 2 x − 4 ⓑ y = 6 x 2 + 7 x − 9 y = 6 x 2 + 7 x − 9

Try It 10.87

ⓐ y = 2 x 2 + 5 x − 2 y = 2 x 2 + 5 x − 2 ⓑ y = −3 x 2 − 4 x + 7 y = −3 x 2 − 4 x + 7

Try It 10.88

ⓐ y = −2 x 2 − 2 x − 3 y = −2 x 2 − 2 x − 3 ⓑ y = 5 x 2 − 2 x − 1 y = 5 x 2 − 2 x − 1

Find the Axis of Symmetry and Vertex of a Parabola

Look again at Figure 10.3 . Do you see that we could fold each parabola in half and that one side would lie on top of the other? The ‘fold line’ is a line of symmetry. We call it the axis of symmetry of the parabola.

We show the same two graphs again with the axis of symmetry in blue. See Figure 10.4 .

The equation of the axis of symmetry can be derived by using the Quadratic Formula. We will omit the derivation here and proceed directly to using the result. The equation of the axis of symmetry of the graph of y = a x 2 + b x + c y = a x 2 + b x + c is x = − b 2 a . x = − b 2 a .

So, to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula x = − b 2 a x = − b 2 a .

Look back at Figure 10.4 . Are these the equations of the dashed red lines?

The point on the parabola that is on the axis of symmetry is the lowest or highest point on the parabola, depending on whether the parabola opens upwards or downwards. This point is called the vertex of the parabola.

We can easily find the coordinates of the vertex, because we know it is on the axis of symmetry. This means its x -coordinate is − b 2 a − b 2 a . To find the y -coordinate of the vertex, we substitute the value of the x -coordinate into the quadratic equation.

Axis of Symmetry and Vertex of a Parabola

For a parabola with equation y = a x 2 + b x + c y = a x 2 + b x + c :

  • The axis of symmetry of a parabola is the line x = − b 2 a x = − b 2 a .
  • The vertex is on the axis of symmetry, so its x -coordinate is − b 2 a − b 2 a .

To find the y -coordinate of the vertex, we substitute x = − b 2 a x = − b 2 a into the quadratic equation.

Example 10.45

For the parabola y = 3 x 2 − 6 x + 2 y = 3 x 2 − 6 x + 2 find: ⓐ the axis of symmetry and ⓑ the vertex.

Try It 10.89

For the parabola y = 2 x 2 − 8 x + 1 y = 2 x 2 − 8 x + 1 find: ⓐ the axis of symmetry and ⓑ the vertex.

Try It 10.90

For the parabola y = 2 x 2 − 4 x − 3 y = 2 x 2 − 4 x − 3 find: ⓐ the axis of symmetry and ⓑ the vertex.

Find the Intercepts of a Parabola

When we graphed linear equations, we often used the x - and y -intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.

Remember, at the y -intercept the value of x x is zero. So, to find the y -intercept, we substitute x = 0 x = 0 into the equation.

Let’s find the y -intercepts of the two parabolas shown in the figure below.

At an x -intercept , the value of y y is zero. To find an x -intercept, we substitute y = 0 y = 0 into the equation. In other words, we will need to solve the equation 0 = a x 2 + b x + c 0 = a x 2 + b x + c for x x .

But solving quadratic equations like this is exactly what we have done earlier in this chapter.

We can now find the x -intercepts of the two parabolas shown in Figure 10.5 .

First, we will find the x -intercepts of a parabola with equation y = x 2 + 4 x + 3 y = x 2 + 4 x + 3 .

Now, we will find the x -intercepts of the parabola with equation y = − x 2 + 4 x + 3 y = − x 2 + 4 x + 3 .

We will use the decimal approximations of the x-intercepts, so that we can locate these points on the graph.

Do these results agree with our graphs? See Figure 10.6 .

Find the intercepts of a parabola.

To find the intercepts of a parabola with equation y = a x 2 + b x + c y = a x 2 + b x + c :

Example 10.46

Find the intercepts of the parabola y = x 2 − 2 x − 8 y = x 2 − 2 x − 8 .

When y = 0 y = 0 , then x = 4 or x = −2 x = 4 or x = −2 . The x -intercepts are the points ( 4 , 0 ) ( 4 , 0 ) and ( −2 , 0 ) ( −2 , 0 ) .

Try It 10.91

Find the intercepts of the parabola y = x 2 + 2 x − 8 . y = x 2 + 2 x − 8 .

Try It 10.92

Find the intercepts of the parabola y = x 2 − 4 x − 12 . y = x 2 − 4 x − 12 .

In this chapter, we have been solving quadratic equations of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 . We solved for x x and the results were the solutions to the equation.

We are now looking at quadratic equations in two variables of the form y = a x 2 + b x + c y = a x 2 + b x + c . The graphs of these equations are parabolas. The x -intercepts of the parabolas occur where y = 0 y = 0 .

For example:

The solutions of the quadratic equation are the x x values of the x -intercepts.

Earlier, we saw that quadratic equations have 2, 1, or 0 solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the equations give the x -intercepts of the graphs, the number of x -intercepts is the same as the number of solutions.

Previously, we used the discriminant to determine the number of solutions of a quadratic equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 . Now, we can use the discriminant to tell us how many x -intercepts there are on the graph.

Before you start solving the quadratic equation to find the values of the x -intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.

Example 10.47

Find the intercepts of the parabola y = 5 x 2 + x + 4 y = 5 x 2 + x + 4 .

Try It 10.93

Find the intercepts of the parabola y = 3 x 2 + 4 x + 4 . y = 3 x 2 + 4 x + 4 .

Try It 10.94

Find the intercepts of the parabola y = x 2 − 4 x − 5 . y = x 2 − 4 x − 5 .

Example 10.48

Find the intercepts of the parabola y = 4 x 2 − 12 x + 9 y = 4 x 2 − 12 x + 9 .

Try It 10.95

Find the intercepts of the parabola y = − x 2 − 12 x − 36 . y = − x 2 − 12 x − 36 .

Try It 10.96

Find the intercepts of the parabola y = 9 x 2 + 12 x + 4 . y = 9 x 2 + 12 x + 4 .

Graph Quadratic Equations in Two Variables

Now, we have all the pieces we need in order to graph a quadratic equation in two variables. We just need to put them together. In the next example, we will see how to do this.

Example 10.49

How to graph a quadratic equation in two variables.

Graph y = x 2 − 6 x + 8 y = x 2 − 6 x + 8 .

Try It 10.97

Graph the parabola y = x 2 + 2 x − 8 . y = x 2 + 2 x − 8 .

Try It 10.98

Graph the parabola y = x 2 − 8 x + 12 . y = x 2 − 8 x + 12 .

Graph a quadratic equation in two variables.

  • Step 1. Write the quadratic equation with y y on one side.
  • Step 2. Determine whether the parabola opens upward or downward.
  • Step 3. Find the axis of symmetry.
  • Step 4. Find the vertex.
  • Step 5. Find the y -intercept. Find the point symmetric to the y -intercept across the axis of symmetry.
  • Step 6. Find the x -intercepts.
  • Step 7. Graph the parabola.

We were able to find the x -intercepts in the last example by factoring. We find the x -intercepts in the next example by factoring, too.

Example 10.50

Graph y = − x 2 + 6 x − 9 y = − x 2 + 6 x − 9 .

Try It 10.99

Graph the parabola y = −3 x 2 + 12 x − 12 . y = −3 x 2 + 12 x − 12 .

Try It 10.100

Graph the parabola y = 25 x 2 + 10 x + 1 . y = 25 x 2 + 10 x + 1 .

For the graph of y = − x 2 + 6 x − 9 y = − x 2 + 6 x − 9 , the vertex and the x -intercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation 0 = − x 2 + 6 x − 9 0 = − x 2 + 6 x − 9 is 0, so there is only one solution. That means there is only one x -intercept, and it is the vertex of the parabola.

How many x -intercepts would you expect to see on the graph of y = x 2 + 4 x + 5 y = x 2 + 4 x + 5 ?

Example 10.51

Graph y = x 2 + 4 x + 5 y = x 2 + 4 x + 5 .

Try It 10.101

Graph the parabola y = 2 x 2 − 6 x + 5 . y = 2 x 2 − 6 x + 5 .

Try It 10.102

Graph the parabola y = −2 x 2 − 1 . y = −2 x 2 − 1 .

Finding the y -intercept by substituting x = 0 x = 0 into the equation is easy, isn’t it? But we needed to use the Quadratic Formula to find the x -intercepts in Example 10.51 . We will use the Quadratic Formula again in the next example.

Example 10.52

Graph y = 2 x 2 − 4 x − 3 y = 2 x 2 − 4 x − 3 .

Try It 10.103

Graph the parabola y = 5 x 2 + 10 x + 3 . y = 5 x 2 + 10 x + 3 .

Try It 10.104

Graph the parabola y = −3 x 2 − 6 x + 5 . y = −3 x 2 − 6 x + 5 .

Solve Maximum and Minimum Applications

Knowing that the vertex of a parabola is the lowest or highest point of the parabola gives us an easy way to determine the minimum or maximum value of a quadratic equation. The y -coordinate of the vertex is the minimum y -value of a parabola that opens upward. It is the maximum y -value of a parabola that opens downward. See Figure 10.7 .

Minimum or Maximum Values of a Quadratic Equation

The y -coordinate of the vertex of the graph of a quadratic equation is the

  • minimum value of the quadratic equation if the parabola opens upward.
  • maximum value of the quadratic equation if the parabola opens downward.

Example 10.53

Find the minimum value of the quadratic equation y = x 2 + 2 x − 8 y = x 2 + 2 x − 8 .

Try It 10.105

Find the maximum or minimum value of the quadratic equation y = x 2 − 8 x + 12 y = x 2 − 8 x + 12 .

Try It 10.106

Find the maximum or minimum value of the quadratic equation y = −4 x 2 + 16 x − 11 y = −4 x 2 + 16 x − 11 .

We have used the formula

to calculate the height in feet, h h , of an object shot upwards into the air with initial velocity, v 0 v 0 , after t t seconds.

This formula is a quadratic equation in the variable t t , so its graph is a parabola. By solving for the coordinates of the vertex, we can find how long it will take the object to reach its maximum height. Then, we can calculate the maximum height.

Example 10.54

The quadratic equation h = −16 t 2 + v 0 t + h 0 h = −16 t 2 + v 0 t + h 0 models the height of a volleyball hit straight upwards with velocity 176 feet per second from a height of 4 feet.

  • ⓐ How many seconds will it take the volleyball to reach its maximum height?
  • ⓑ Find the maximum height of the volleyball.

h = −16 t 2 + 176 t + 4 h = −16 t 2 + 176 t + 4

Since a is negative, the parabola opens downward.

The quadratic equation has a maximum.

  • ⓐ Find the axis of symmetry. t = − b 2 a t = − 176 2 ( −16 ) t = 5.5 The axis of symmetry is t = 5.5 . The vertex is on the line t = 5.5 . The maximum occurs when t = 5.5 seconds. Find the axis of symmetry. t = − b 2 a t = − 176 2 ( −16 ) t = 5.5 The axis of symmetry is t = 5.5 . The vertex is on the line t = 5.5 . The maximum occurs when t = 5.5 seconds.

Try It 10.107

The quadratic equation h = −16 t 2 + 128 t + 32 h = −16 t 2 + 128 t + 32 is used to find the height of a stone thrown upward from a height of 32 feet at a rate of 128 ft/sec. How long will it take for the stone to reach its maximum height? What is the maximum height? Round answers to the nearest tenth.

Try It 10.108

A toy rocket shot upward from the ground at a rate of 208 ft/sec has the quadratic equation of h = −16 t 2 + 208 t h = −16 t 2 + 208 t . When will the rocket reach its maximum height? What will be the maximum height? Round answers to the nearest tenth.

Access these online resources for additional instruction and practice graphing quadratic equations:

  • Graphing Quadratic Functions
  • How do you graph a quadratic function?
  • Graphing Quadratic Equations

Section 10.5 Exercises

Practice makes perfect.

In the following exercises, graph:

y = x 2 + 3 y = x 2 + 3

y = − x 2 + 1 y = − x 2 + 1

In the following exercises, determine if the parabola opens up or down.

y = −2 x 2 − 6 x − 7 y = −2 x 2 − 6 x − 7

y = 6 x 2 + 2 x + 3 y = 6 x 2 + 2 x + 3

y = 4 x 2 + x − 4 y = 4 x 2 + x − 4

y = −9 x 2 − 24 x − 16 y = −9 x 2 − 24 x − 16

In the following exercises, find ⓐ the axis of symmetry and ⓑ the vertex.

y = x 2 + 8 x − 1 y = x 2 + 8 x − 1

y = x 2 + 10 x + 25 y = x 2 + 10 x + 25

y = − x 2 + 2 x + 5 y = − x 2 + 2 x + 5

y = −2 x 2 − 8 x − 3 y = −2 x 2 − 8 x − 3

In the following exercises, find the x - and y -intercepts.

y = x 2 + 7 x + 6 y = x 2 + 7 x + 6

y = x 2 + 10 x − 11 y = x 2 + 10 x − 11

y = − x 2 + 8 x − 19 y = − x 2 + 8 x − 19

y = x 2 + 6 x + 13 y = x 2 + 6 x + 13

y = 4 x 2 − 20 x + 25 y = 4 x 2 − 20 x + 25

y = − x 2 − 14 x − 49 y = − x 2 − 14 x − 49

In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry.

y = x 2 + 6 x + 5 y = x 2 + 6 x + 5

y = x 2 + 4 x − 12 y = x 2 + 4 x − 12

y = x 2 + 4 x + 3 y = x 2 + 4 x + 3

y = x 2 − 6 x + 8 y = x 2 − 6 x + 8

y = 9 x 2 + 12 x + 4 y = 9 x 2 + 12 x + 4

y = − x 2 + 8 x − 16 y = − x 2 + 8 x − 16

y = − x 2 + 2 x − 7 y = − x 2 + 2 x − 7

y = 5 x 2 + 2 y = 5 x 2 + 2

y = 2 x 2 − 4 x + 1 y = 2 x 2 − 4 x + 1

y = 3 x 2 − 6 x − 1 y = 3 x 2 − 6 x − 1

y = 2 x 2 − 4 x + 2 y = 2 x 2 − 4 x + 2

y = −4 x 2 − 6 x − 2 y = −4 x 2 − 6 x − 2

y = − x 2 − 4 x + 2 y = − x 2 − 4 x + 2

y = x 2 + 6 x + 8 y = x 2 + 6 x + 8

y = 5 x 2 − 10 x + 8 y = 5 x 2 − 10 x + 8

y = −16 x 2 + 24 x − 9 y = −16 x 2 + 24 x − 9

y = 3 x 2 + 18 x + 20 y = 3 x 2 + 18 x + 20

y = −2 x 2 + 8 x − 10 y = −2 x 2 + 8 x − 10

In the following exercises, find the maximum or minimum value.

y = 2 x 2 + x − 1 y = 2 x 2 + x − 1

y = −4 x 2 + 12 x − 5 y = −4 x 2 + 12 x − 5

y = x 2 − 6 x + 15 y = x 2 − 6 x + 15

y = − x 2 + 4 x − 5 y = − x 2 + 4 x − 5

y = −9 x 2 + 16 y = −9 x 2 + 16

y = 4 x 2 − 49 y = 4 x 2 − 49

In the following exercises, solve. Round answers to the nearest tenth.

An arrow is shot vertically upward from a platform 45 feet high at a rate of 168 ft/sec. Use the quadratic equation h = −16 t 2 + 168 t + 45 h = −16 t 2 + 168 t + 45 to find how long it will take the arrow to reach its maximum height, and then find the maximum height.

A stone is thrown vertically upward from a platform that is 20 feet high at a rate of 160 ft/sec. Use the quadratic equation h = −16 t 2 + 160 t + 20 h = −16 t 2 + 160 t + 20 to find how long it will take the stone to reach its maximum height, and then find the maximum height.

A computer store owner estimates that by charging x x dollars each for a certain computer, he can sell 40 − x 40 − x computers each week. The quadratic equation R = − x 2 + 40 x R = − x 2 + 40 x is used to find the revenue, R R , received when the selling price of a computer is x x . Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.

A retailer who sells backpacks estimates that, by selling them for x x dollars each, he will be able to sell 100 − x 100 − x backpacks a month. The quadratic equation R = − x 2 + 100 x R = − x 2 + 100 x is used to find the R R received when the selling price of a backpack is x x . Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.

A rancher is going to fence three sides of a corral next to a river. He needs to maximize the corral area using 240 feet of fencing. The quadratic equation A = x ( 240 − 2 x ) A = x ( 240 − 2 x ) gives the area of the corral, A A , for the length, x , x , of the corral along the river. Find the length of the corral along the river that will give the maximum area, and then find the maximum area of the corral.

A veterinarian is enclosing a rectangular outdoor running area against his building for the dogs he cares for. He needs to maximize the area using 100 feet of fencing. The quadratic equation A = x ( 100 − 2 x ) A = x ( 100 − 2 x ) gives the area, A A , of the dog run for the length, x x , of the building that will border the dog run. Find the length of the building that should border the dog run to give the maximum area, and then find the maximum area of the dog run.

Everyday Math

In the previous set of exercises, you worked with the quadratic equation R = − x 2 + 40 x R = − x 2 + 40 x that modeled the revenue received from selling computers at a price of x x dollars. You found the selling price that would give the maximum revenue and calculated the maximum revenue. Now you will look at more characteristics of this model. ⓐ Graph the equation R = − x 2 + 40 x R = − x 2 + 40 x . ⓑ Find the values of the x -intercepts.

In the previous set of exercises, you worked with the quadratic equation R = − x 2 + 100 x R = − x 2 + 100 x that modeled the revenue received from selling backpacks at a price of x x dollars. You found the selling price that would give the maximum revenue and calculated the maximum revenue. Now you will look at more characteristics of this model. ⓐ Graph the equation R = − x 2 + 100 x R = − x 2 + 100 x . ⓑ Find the values of the x -intercepts.

Writing Exercises

For the revenue model in Exercise 10.205 and Exercise 10.209 , explain what the x -intercepts mean to the computer store owner.

For the revenue model in Exercise 10.206 and Exercise 10.210 , explain what the x -intercepts mean to the backpack retailer.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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  • Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis
  • Publisher/website: OpenStax
  • Book title: Elementary Algebra 2e
  • Publication date: Apr 22, 2020
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/elementary-algebra-2e/pages/1-introduction
  • Section URL: https://openstax.org/books/elementary-algebra-2e/pages/10-5-graphing-quadratic-equations-in-two-variables

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Unit 8 Homework 2 Graphing Quadratic Equations

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2.1.2: Graphs of Quadratic Functions

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Graphs of Quadratic Functions

In prior lessons, we have discussed the standard form of a quadratic function : f(x)=ax 2 +bx+c. You may have seen other forms, such as the vertex form , or the factored form . Why are there so many common ways to write the same equations? Why should we learn all these different forms if the standard form is the most common anyway?

Graphing Quadratic Functions

Quadratic functions.

A function f defined by f(x)=ax 2 +bx+c, where a,b, and c are real numbers and a≠0, is called a quadratic function .

The defining characteristic of a quadratic function is that it is a polynomial whose highest exponent is 2.

There are several ways to write quadratic functions:

  • standard form , the form of the quadratic function above: f(x)=ax 2 +bx+c
  • vertex form , commonly used for quick sketching: f(x)=a(x−h) 2 +k
  • factored form , excellent for finding x -intercepts: f(x)=a(x−r 1 )(x−r 2 )

The y− intercept of a quadratic function in standard form is (0,c) and it is found by substituting for x in f(x)=ax 2 +bx+c.

Summary of Vertex Form

Given a quadratic function in vertex form : f(x)=a(x−h) 2 +k:

  • The vertex is at (h,k)
  • The parabola opens up if a>0
  • The parabola opens down if a<0
  • The parabola is narrower than y=x 2 if |a|>1
  • The parabola opens wider than y=x 2 if |a|<1

Vertex of a Parabola

In the standard form of a quadratic function, the x−coordinate of the vertex of the parabola is given by the equation:

\(\ x=-\frac{b}{2 a}\)

The y−coordinate of the vertex is found with:

\(\ y=f\left(-\frac{b}{2 a}\right)\)

Axis of Symmetry of a Parabola

A parabola has reflective symmetry about a vertical line through the vertex.

The vertical line \(\ x=-\frac{b}{2 a}\) is also the parabola's axis of symmetry .

Graph g(x)=x 2 +6x+7 using transformations .

First, complete the square to write this function in vertex form. Add and subtract \(\ \left(\frac{b}{2}\right)^{2}\) to the right hand side of the equation:

g(x)=x 2 +6x+7

=x 2 +6x+9+7−9

Now, factor the right hand side:

g(x)=(x+3) 2 −2

Thus, a=1 and the vertex of this parabola is (-3, -2). We know that the parabola opens up with the same width as y=x 2 and it has a minimum value at the vertex. The graph of the parabola is below.

f-d_b5355b73c804c84da9989ac299c220edfd8fce277cd15962a7c289d9+IMAGE_TINY+IMAGE_TINY.jpg

Find the vertex and graph the quadratic function g(x)=x 2 −8x+12.

The x−coordinate of the vertex is \(\ x=-\frac{-8}{2}=4\).

The y−coordinate of the vertex is g(4)=(4) 2 −8(4)+12=16−32+12=−4

Thus the vertex is at (4, -4).

To graph the parabola, we will make a table of points staring with the x−coordinate of 4:

f-d_45539f6668cc2d67ed5a73972249d196e68558b02b7591297505153d+IMAGE_TINY+IMAGE_TINY.jpg

Sketch the graph of the function y=f(x)=x 2 +2x−3.

Let's first find the intercepts. For the y−intercept, if x=0, then f(0)=−3, or y=−3, so the y−intercept point is (0, -3).

Now, for the x−intercepts, if y=f(x)=0, then x 2 +2x−3=0, or x 2 +2x−3=(x+3)(x−1)=0

so that x=−3 and x=1 are the x−intercepts, that is (-3, 0) and (1, 0).

The vertex (extreme point) is at

\(\ x=\frac{-b}{2 a}=\frac{-2}{2(1)}=-1\)

\(\ \begin{aligned} f(-1) &=(-1)^{2}+2(-1)-3 \\ &=-4 \end{aligned}\)

The vertex is (-1, -4).

Since the coefficient of x 2 is positive, a>0, the extreme point is a minimum and the parabola opens up. From this information, we can make a rough sketch of the parabola containing the points determined above. Notice that the range of the function is y≥−4.

f-d_883851f2958f4319065de5b55c3fd00494069798f628274ded2901be+IMAGE_TINY+IMAGE_TINY.jpg

Sketch the graph of the quadratic function f(x)=−x 2 +4x.

To find the y−intercept, set x=0, and f(0)=−(0) 2 +4(0)=0

Thus the parabola intercepts the y−axis at the origin.

The x−intercept is obtained by setting y=0. Thus, −x 2 +4x=0.

−x 2 +4x=−x(x−4)=0

so that x=0 and x=4 are the x−intercepts.

We have a=−1 and b=4, so that the extreme point occurs when

\(\ x=\frac{-b}{2 a}=\frac{-4}{2(-1)}=2\)

Since f(2)=−(2) 2 +4(2)=−4+8=4, then (2, 4) is the extreme point. It is a maximum point since a=−1<0 and the parabola opens down. Finally, the graph can be obtained by sketching a parabola through the points determined above. From graph, the range of the function is y≤4.

f-d_5175b6dee29a23369848e34ac48458c519de4b7d087142019d3a3e96+IMAGE_TINY+IMAGE_TINY.jpg

Sketch the graph of y=−3(x−2) 2 +1.

The equation y=−3(x−2) 2 +1 is already in vertex form, so graphing is relatively easy:

Recall the when the equation is written in vertex form like this, the vertex is the point (h,k):

vertex form: y=a(x−h) 2 +k

Our equation: y=−3(x−2) 2 +1

Examining our equation, we can see the vertex of the parabola is at (2, 1).

To find another point on the parabola solve for an x value.

Since the vertex is at x = 2, let's try one unit to the right: x = 3.

−3(3−2) 2 +1=−2

∴ There is a point on the parabola at (3, -2)

Since a parabola has an axis of symmetry that passes through its vertex, we can reflect the point (3, -2) across the axis of symmetry to get another point, (1, -2) also on the parabola.

The graph of y=−3(x−2)2+1 is shown below, using vertex (2, 1) and points (3, -2) and (1, -2).

  • What is the U-Shaped graph of a quadratic function called?
  • Which direction does a parabola open if the leading coefficient ( a ) is positive?
  • For y 2 =x If the coefficient of y is positive, which way does the parabola open?
  • What is the name of the lowest point of a parabola that opens up and the highest point of a parabola that opens down.
  • What is the name of the line passing through the vertex that divides the parabola into two symmetric parts?
  • Sketch the graph of y=x 2 +3
  • Sketch the graph of y=−x 2 +4x−4
  • Sketch the graph of y=2x 2 +8x
  • Consider the following quadratic function: y=−x 2 −2x+1 a) Which direction does it open? b) What is the vertex? c) Is it stretched in any way?
  • Consider the quadratic functions: y=2x 2 y=4x 2 y=6x 2 Which quadratic function would you expect to have the narrowest parabola? Explain your answer.

Sketch the graph of each function:

  • y=−x 2
  • y=3x 2 +6x+1
  • y=12x 2 +2x+4
  • y=(x−3) 2 +4
  • y=−x 2 −8x−17

The quadratic function y=−0.05x 2 +1.5x can be used to represent the path of a football kicked 30 yards down the field. The variable x represents the distance, in yards, the ball has traveled down the field. The height, in yards, of the football in the air is represented by the variable (y).

Use the quadratic function to calculate the height of the ball as it travels down the field. Round your answers to the nearest hundredth of a yard.

  • What is the maximum height of the football during the kick?
  • How far down the field has the football traveled when it reaches its maximum height?
  • Use the information in the table to graph the path of the football kick.
  • If you were shown only the graph of this quadratic function, how could you determine the maximum height of the football during the kick and how far down the field the football has traveled when it reaches its maximum height?

Review (Answers)

To see the Review answers, open this PDF file and look for section 2.2.

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Graphing Quadratic Equations - Homework 2

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NEW CHAPTER 10 Quadratic Equations

10.2: graphs of quadratic equations, learning outcomes.

  • Graph quadratic equations of the form [latex]y=ax^2+bx+c[/latex]
  • Identify important features of the graphs of quadratic equations
  • Determine the maximum or minimum value of a quadratic equation
  • Parent equation : the simplest form of a general equation
  • Parabola : the shape of any quadratic equation
  • Vertex : the turning point of a parabola
  • Line of symmetry : a line that cuts the graph into two mirror images

Graphing Quadratic Equations Using Tables

The simplest form of a quadratic equation is [latex]y=ax^2[/latex]. This is also referred to as the  parent equation of any quadratic equation. The basic shape of a quadratic equation is a parabola . It has a vertex where the parabola turns and a line of symmetry that runs through the parabola and splits the graph into two mirror images. We discovered all of this in the last section by determining solutions of the equation and plotting the solution points. We can use this technique for any quadratic equation.

Complete a table of values for the equation [latex]y=2x^2[/latex], then graph the equation.

To create a table of values, we can choose any [latex]x[/latex]-values and find the corresponding [latex]y[/latex]-values.

To create the graph, we plot the points and join the dots.

y=2x^2 with points

Notice that it is still a parabola. All quadratic equations take the shape of a parabola.

Complete a table of values for the equation [latex]y=-x^2+4x-3[/latex], then graph the equation. State the vertex and axis of symmetry.

To create the graph, we plot the points.

Plot of first points

Unfortunately, the points we chose do not show any symmetry or a turning point. However, the graph looks like it will turn to the right of [latex]x=2[/latex], so let’s find a few other points that lie to the right of [latex]x=2[/latex].

Now we have some symmetry and can join the dots to create the graph.

Graph of y=-x^2+4x-3

The vertex is at [latex](2, 1)[/latex] and the axis of symmetry is the vertical line [latex]x=2[/latex].

Notice also, that the point [latex](-2, -15)[/latex] has a twin as a mirror image at [latex](6, -15)[/latex].

In the first example, the parabola opens upwards, while in the second example, the parabola opens downwards. This is determined by the value of [latex]a[/latex] in the equation [latex]y=ax^2+bx+c[/latex]. When [latex]a>0[/latex], the parabola opens upwards. When [latex]a<0[/latex], the parabola opens downwards.  Notice that [latex]a\neq0[/latex] because that would turn the quadratic equation [latex]y=ax^2+bx+c[/latex] into a linear equation [latex]y=bx+c[/latex].

Complete a table of values for the equation [latex]y=-x^2+7[/latex], then graph the equation. State the vertex and axis of symmetry.

First notice that [latex]a=-1[/latex] so the parabola will open downwards.

To complete a table of values we can choose any [latex]x[/latex]-values:

y=-x^2+7 with points

From the graph the vertex is at [latex](0, 7)[/latex] and the line of symmetry is the vertical line [latex]x=0[/latex].

Sometimes it can be messy to draw a graph using just solutions. Deciding which [latex]x[/latex]-values to choose can be bothersome. It can also be impossible to determine exactly where the axis of symmetry and vertex lie if they do not contain integer values. That’s why we use other features of the graph to help us.  For example, it would be helpful to know exactly where the vertex and axis of symmetry lie. Or, where the graph crosses the axes. So, let’s discover how to determine such features.

Features of Parabolas

Intercepts: the [latex]y[/latex]-intercept.

The [latex]y[/latex]-intercept of any graph is found by setting [latex]x=0[/latex] in the equation of the graph and solving for [latex]y[/latex]. For a parabola with equation [latex]y=ax^2+bx+c[/latex], setting [latex]x=0[/latex] results in [latex]y=c[/latex]. Consequently, the [latex]y[/latex]-intercept of any parabola is always the point [latex](0, c)[/latex].

Determine the [latex]y[/latex]-intercept of the graph with equation:

1. [latex]y=4x^2-3x+6[/latex]

2. [latex]y=-2x^2-7[/latex]

3. [latex]y=x^2+5x[/latex]

1. Since [latex]c=6[/latex], the [latex]y[/latex]-intercept is the point [latex](0, 6)[/latex].

2. Since [latex]c=-7[/latex], the [latex]y[/latex]-intercept is the point [latex](0, -7)[/latex].

3. Since [latex]c=0[/latex], the [latex]y[/latex]-intercept is the point [latex](0, 0)[/latex].

1. [latex]y=x^2-4x+1[/latex]

2. [latex]y=-2x^2-3[/latex]

3. [latex]y=x^2-6x[/latex]

  • [latex](0, 1)[/latex]
  • [latex](0, -3)[/latex]
  • [latex](0, 0)[/latex]

Intercepts: The [latex]x[/latex]-Intercepts

The [latex]x[/latex]-intercepts of a parabola occur where the parabola crosses the [latex]x[/latex]-axis (figure 1). Specifically, this is where [latex]y=0[/latex]. So to find the [latex]x[/latex]-intercepts of a parabola, we must solve the equation [latex]ax^2+bx+c=0[/latex].

y=x^2-x-2

Figure 1. [latex]x[/latex]-intercepts

Provided this quadratic equation factors, we can solve it using the zero-factor property.  If the equation does not factor, there are alternative ways of solving it that will be taught in the next course.

Find the [latex]x[/latex]-intercepts for the graph of [latex]y={x}^{2}-x-6[/latex].

The [latex]x[/latex]-intercepts are found by solving the equation [latex]x^2 - x - 6 = 0[/latex].

The [latex]x[/latex]-intercepts are [latex]\left(3, 0 \right)[/latex] and [latex] \left( -2, 0 \right) [/latex].

Find the [latex]x[/latex]-intercepts for the graph of [latex]y={x}^{2}-3x+2[/latex].

The graph has two [latex]x[/latex]-intercepts at [latex](1, 0)[/latex] and [latex](2, 0)[/latex]

Find the [latex]x[/latex]-intercepts for the graph of [latex]y={x}^{2}-6x+9[/latex].

The graph has one [latex]x[/latex]-intercept at [latex](3, 0)[/latex].

Parabolas whose equations factor have [latex]x[/latex]-intercepts. But not all parabolas have [latex]x[/latex]-intercepts. Consider the parabolas in figure 2:

Parabolas and x-intercepts

Figure 2. Parabolas with 2, 1, or 0 [latex]x[/latex]-intercepts.

Parabolas can have two [latex]x[/latex]-intercepts, one [latex]x[/latex]-intercept, or no [latex]x[/latex]-intercepts.

Find the [latex]x[/latex]-intercepts of the graph of [latex]y = x^2 + 4[/latex].

To find the [latex]x[/latex]-intercepts, we need to solve the equation [latex]x^2 + 4 = 0[/latex].

However, [latex]x^2 + 4[/latex] does not factor.

We can solve this equation by rewriting it:

[latex]x^2 + 4=0[/latex] can be rewritten as [latex]x^2 = -4 [/latex].

There are no real number values for [latex]x[/latex] that when squared result in [latex]-4[/latex].

The only value that [latex]x[/latex] can be are the imaginary numbers [latex]2i[/latex] and [latex]-2i[/latex] because [latex](2i)^2=4i^2=4(-1)=-4[/latex] and [latex](-2i)^2=4i^2=4(-1)=-4[/latex].

Since [latex]2i[/latex] and [latex]-2i[/latex] are complex numbers, they will not show up on the graph.

This parabola will have no [latex]x[/latex]-intercepts.

If a parabola does not intersect with the [latex]x[/latex]-axis,and therefore has no [latex]x[/latex]-intercepts, there are complex number solutions to the equation [latex]y=ax^2+bx+c=0[/latex]. Such solutions cannot be graphed on the real number line, so will not appear on the graph.

Find the [latex]x[/latex]-intercepts of the graph of [latex]y = 9x^2 + 1[/latex].

The graph has no [latex]x[/latex]-intercepts.

Axis of Symmetry and the Vertex

The axis of symmetry and the vertex are very important features of a parabola so being able to find them will be extremely useful for graphing.

Axis of SYmmetry and vertex

For the graph of the equation [latex]y=ax^2+bx+c[/latex], the axis of symmetry is the vertical line [latex]x=-\frac{b}{2a}[/latex].

The vertex is the point where [latex]x=-\frac{b}{2a}[/latex], paired with the corresponding [latex]y[/latex]-value.

For example, consider the equation [latex]y=2x^2-3x+4[/latex]. To find the axis of symmetry calculate [latex]x=-\frac{b}{2a}=-\frac{-3}{2\cdot 2}=\frac{3}{4}[/latex]. The axis of symmetry is the vertical line with equation [latex]x=\frac{3}{4}[/latex].

The axis of symmetry passes through the vertex, so the [latex]x[/latex]-coordinate of the vertex is also [latex]\frac{3}{4}[/latex]. To find the [latex]y[/latex]-coordinate, we substitute [latex]x=\frac{3}{4}[/latex] into the original equation [latex]y=2x^2-3x+4[/latex]:

[latex]\begin{equation}\begin{aligned}y&=2x^2-3x+4 \\ y&=2 {\left (\frac{3}{4}\right )}^{2}-3\cdot\frac{3}{4}+4 \\ y&=2\cdot\frac{9}{16}-\frac{9}{4}+4\\y&=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\y&=\frac{23}{8}\end{aligned}\end{equation}[/latex]

The vertex is at the point [latex]\left(\dfrac{3}{4},\dfrac{23}{8}\right)[/latex]. This is verified by the graph in figure 3:

graph of y equal two x squared -3x+4

Figure 3. Graph of [latex]y=2x^2-3x+4[/latex]

Find the axis of symmetry and the vertex of the graph with equation [latex]y=-3x^2+x-2[/latex].

Axis:  [latex]x=-\frac{b}{2a}=-\frac{1}{2\cdot (-3)}=-\frac{1}{-6}=\frac{1}{6}[/latex]

Vertex: [latex]x=\frac{1}{6}[/latex], so

[latex]\begin{equation}\begin{aligned}y&=-3x^2+x-2 \\ y&=-3{\left (\frac{1}{6}\right )}^{2}+\frac{1}{6}-2 \\ y&=-\frac{3}{1}\frac{1}{36}+\frac{1}{6}-2 \\ y&=-\frac{1}{12}+\frac{1}{6}-2 \\ y&=\frac{-1+2-24}{12} \\ &y=-\frac{23}{12}\end{aligned}\end{equation}[/latex]

Graph of y=-3x^2+x-2

Find the axis of symmetry and the vertex of the graph of the equation [latex]y=2x^2+2x-4[/latex].

Axis: [latex]x=\frac{1}{2}[/latex]

Vertex: [latex]\left (\frac{1}{2}, -\frac{9}{2}\right )[/latex]

Putting it all Together

We have learned some important things about the graphs of quadratic equations that will make it easier for us to create a graph.  These features of a parabola are summarized below:

Graphs of Quadratic Functions

For [latex] \displaystyle y=a{{x}^{2}}+bx+c[/latex], where [latex]a, b[/latex] and [latex]c[/latex] are real numbers, and [latex]a\neq0[/latex],

  • The parabola opens upward if [latex]a > 0[/latex] and downward if [latex]a < 0[/latex].
  • The [latex]y[/latex]-intercept of the parabola occurs at the point [latex](0,c)[/latex].
  • The [latex]x[/latex]-intercepts are found by solving the equation [latex]ax^2+bx+c=0[/latex]. The [latex]y[/latex]-coordinate is zero. There may be 0, 1, or 2 [latex]x[/latex]-intercepts.
  • The axis of symmetry is the vertical line [latex]x=\frac{-b}{2a}[/latex].
  • The vertex has an [latex]x[/latex]-coordinate of [latex]x=\dfrac{-b}{2a}[/latex]. The [latex]y[/latex]-coordinate is found by substituting this [latex]x[/latex]-value into the equation and solving for [latex]y[/latex].

We can use the properties of parabolas to help us graph a quadratic equation of the form [latex]y=ax^2+bx+c[/latex] without having to calculate an exhaustive table of values.

Graph [latex]y=−2x^{2}+3x–3[/latex].

Let’s start by considering the features of a parabola.

[latex]a=-2[/latex] so the parabola opens downwards.

Since [latex]|a|>1[/latex] the graph will be narrower than the graph of [latex]y=x^2[/latex].

[latex]y[/latex]-intercept :

Since [latex]c=-3[/latex], the [latex]y[/latex]-intercept will be [latex](0, -3)[/latex].

Axis of symmetry:

[latex]x=\frac{-b}{2a}=\frac{-3}{2\cdot (-2)}=\frac{3}{4}[/latex].

[latex]x=\frac{3}{4}[/latex], so [latex]y=−2x^{2}+3x–3=-2{\left (\frac{3}{4}\right )}^{2}+3\cdot \frac{3}{4}-3=-2\cdot \frac{9}{16}+\frac{9}{4}-3=\frac{-9}{8}+\frac{9}{4}-3=\frac{-9+18-24}{8}=\frac{-15}{8}[/latex].  The vertex is the point [latex]\left (\frac{3}{4}, -\frac{15}{8}\right )[/latex].

[latex]x[/latex]-intercepts : 

[latex]−2x^{2}+3x–3[/latex] does not factor, so we cannot find the [latex]x[/latex]-intercepts .

Let’s graph everything we have:

1st draft of graph

This twin has the same [latex]y[/latex]-value of [latex]-3[/latex]. To find the [latex]x[/latex]-value, move the same number of units past the axis of symmetry to the right.

The point [latex](0, -3)[/latex] is [latex]\frac{3}{4}[/latex] of a unit to the left of the axis, so its twin is [latex]\frac{3}{4}[/latex] of a unit to the right of the axis at [latex]\left (\frac{3}{2}, -3\right )[/latex].

From the information we have found so far, we have the turning point and direction of the graph. Now let’s find some solutions and use their twins to plot more points.

2nd draft

Finally connect the points as best you can using a smooth curve.

Determine the maximum value of [latex]y[/latex]: [latex]y=-4x^2-4x+3[/latex]

Since the graph is a parabola that opens downwards, the maximum value will occur at the vertex.

Find the vertex:  [latex]x=\frac{-b}{2a}=\frac{4}{2\cdot (-4)}=\frac{-1}{2}[/latex]

Then find the [latex]y[/latex]-value:

[latex]y=-4x^2-4x+3\\y=-4{\left (\frac{-1}{2}\right)}^{2}-4\left (\frac{-1}{2}\right )+3\\y=\frac{-4}{1}\cdot\frac{1}{4}+2+3\\y=-1+2+3\\y=4[/latex]

The maximum value is [latex]y=4[/latex].

Determine the minimum value of [latex]y[/latex]:  [latex]y=x^2-8x-4[/latex]

The minimum value is [latex]-20[/latex]

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About Graphing Quadratic Functions

Quadratic function has the form $ f(x) = ax^2 + bx + c $ where a, b and c are numbers

You can sketch quadratic function in 4 steps. I will explain these steps in following examples.

Sketch the graph of the quadratic function

In this case we have $ a=1, b=2 $ and $c=-3$

STEP 1: Find the vertex.

To find x - coordinate of the vertex we use formula:

So, we substitute $1$ in for $a$ and $2$ in for $b$ to get

To find y - coordinate plug in $x=-1$ into the original equation:

So, the vertex of the parabola is $ {\color{red}{ (-1,-4) }} $

STEP 2: Find the y-intercept.

To find y - intercept plug in $x=0$ into the original equation:

quadratic function graph

So, the y-intercept of the parabola is $ {\color{blue}{ y = -3 }} $

STEP 3: Find the x-intercept.

To find x - intercept solve quadratic equation $f(x)=0$ in our case we have:

Solutions for this equation are:

( to learn how to solve quadratic equation use quadratic equation solver )

STEP 4: plot the parabola.

Here we have $ a=-1, b=2 $ and $c=-2$

The x-coordinate of the vertex is:

quadratic function graph 2

The y-coordinate of the vertex is:

The y-intercept is:

In this case x-intercept doesn't exist since equation $-x^2+2x-2=0$ does not has the solutions (use quadratic equation solver to check ). So, in this case we will plot the graph using only two points

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3.1: Graphs of Quadratic Functions

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Graphs of Quadratic Functions

A Quadratic Function is any function defined by a polynomial whose greatest exponent is two. That means it can be written in the form \(f(x)=ax^2+bx+c\), with the restrictions that the parameters \(a\), \(b\), and \(c\) are real numbers and \(a\) canNOT be zero.  

The graph of any quadratic function is a U-shaped curve called a parabola . There are certain key features that are important to recognize on a graph and to calculate from an equation.

Key features of a parabola

Example \(\PageIndex{1}\): Identify Features of a Parabola from a graph

Determine features of th parabola illustrated below.

Forms of a Quadratic Function

There are two important forms of a quadratic function

Definitions: Forms of Quadratic Functions

A quadratic function is a polynomial function of degree two. The graph of a quadratic function is a parabola.

  • The general form of a quadratic function is \(f(x)=ax^2+bx+c\) with real number parameters \(a\), \(b\), and \(c\) and \(a{\neq}0\).
  • The standard form or vertex form of a quadratic function is \(f(x)=a(x−h)^2+k\) with real number parameters \(a\), \(h\), and \(k\) and \(a{\neq}0\).  NOTICE the minus sign in front of the parameter \(h\)!!

The graph of a quadratic in standard form, \(f(x)=a(x−h)^2+k\), is a graph of  \(y=x^2\) that has been shifted horizontally \(h\) units and vertically \(k\) units. Thus the vertex, originally at at \((0,0)\), is located at the point \((h, k)\) in the graph of \(f\). 

A formula for the location of the vertex for a quadratic in general form can be found by equating the two forms for a quadratic.

\[\begin{align*} a(x−h)^2+k &= ax^2+bx+c \\[4pt] ax^2−2ahx+(ah^2+k)&=ax^2+bx+c \end{align*} \]

This is an identity, so it is true for ALL values of \(x\): the coefficients for the \(x^2\) terms must be the same, the coefficients for the \(x\) terms must be the same, and the constant terms must be the same. Equating the \(x\) coefficients we get

\[–2ah=b \text{, so } h=−\dfrac{b}{2a}. \nonumber\]

This is a formula for the \(x\)-coordinate of the vertex. The \(y\)-coordinate of the vertex is \(y = f(h)=k\).

Features of the graph of a quadratic function depend on the parameter values  \(a\), \(b\), \(c\)  or \(a\), \(h\), \(k\) used in its equation. How features of the parabola for a quadratic function can be obtained is summarized below.

how-to.png

  • \(a > 0\), the parabola opens up  \( \stackrel {+  \:\:\:    +}  {\bigcup} \)
  • \(a < 0 \), the parabola opens down  \( \stackrel{- \:\:\:  -}{\bigcap } \)
  • If the function is in general form, calculate \(h\) and \(k\):  \(h=\dfrac{-b}{2a}, \qquad k=f(h)=f(\dfrac{−b}{2a}).\)
  • The axis of symmetry , \(x=h\) is the \( \underline {\textrm{equation}} \) of the vertical line through the vertex.
  • If (\a > 0\), the parabola opens up and the vertex is the lowest point on the graph with a  minimum value \(k\).
  • If (\a < 0\), the parabola opens down and the vertex is the highest point on the graph with a  maximum value \(k\).
  • If \(a > 0\), the vertex is the lowest point on the graph so the range of the function is \( [k,\infty) \).
  • If \(a < 0\), the vertex is the highest point on the graph so the range of the function is \((-\infty, k] \).
  • The \(x\)-intercepts are the points \((s, 0)\), where \(s\) is a real solution to \(f(x)=0\)
  • The \(y\)-intercept is the point \((0, f(0))\).

Orientation

When the quadratic term, \(ax^2\), is positive, the parabola opens upward, and when the quadratic term is negative, the parabola opens downward.

Example \(\PageIndex{2}\): Find the orientation of a parabola

Determine whether each parabola opens upward or downward:

try-it.png

Determine whether the graph of each function is a parabola that opens upward or downward:

Vertex and Axis of Symmetry

When given a quadratic in standard form \(f(x)=a(x−h)^2+k\), the vertex and axis of symmetry is easily found once the parameters \(h\) and \(k\) have been identified. (Notice the sign on \(h\)!!) The vertex is \((h, k)\) and the axis of symmetry is the vertical line \( x=h\).

When given a quadratic in general form: \(f(x)=ax^2+bx+c\), more computation is required. After identifying parameters \(a\) and \(b\), calculate \(h=–\dfrac{b}{2a}\). Then find the corresponding \(y\) coordinate for that point on the graph: \( y=f(h)=k \). Once \(h\) and \(k\) have been determined, the vertex is at \((h, k)\) and the axis of symmetry is the vertical line \( x=h\).

Example \(\PageIndex{3a}\): Find the Vertex from the General Form of the Quadratic Equation

For the graph of \(f(x)=3x^{2}-6 x+2\) find (a) the axis of symmetry and (b) the vertex.

\( \begin{array}{llc} \text{a.} & \text{Identify the equation parameters} & a=3, b=-6, c=2 \\ & \text{The axis of symmetry is the vertical line } x=-\frac{b}{2 a}  & \\ & \text{Substitute the values }a \text{ and } b \text{ into the formula} & x=-\frac{-6}{2 \cdot 3}=1 \\ && \text{The axis of symmetry is the line } x=1\\ \end{array} \)

\( \begin{array}{llc} \text{b.} & \text{The vertex is a point on the line of symmetry, so} & \text{ The } x \text{ coordinate of the vertex is } x=1\\ & \text{The }y \text{ coordinate will be }f(1) & f(1)=3({\color{red}{1}})^2-6({\color{red}{1}})+2 \\ & \text{Simplify} & f(1) = 3-6+2 \\ & \text{The result is the }y\text{ coordinate of the vertex.} & f(1)=-1 \\ && \text{The vertex is } (1, -1)\\ \end{array} \)

Example \(\PageIndex{3b}\): Find the Vertex from the Standard Form of the Quadratic Equation

For the graph of \(f(x)=6(x-3)^{2}+4\) find:

  • the axis of symmetry

\( \begin{array}{lll} \text{a.} & \text{Identify the equation parameters} & a=6, h=3, k=4 \\ & \text{The axis of symmetry is the vertical line } x=h & \\ & \text{Substitute.} &  \text{The axis of symmetry is the line } x=3\\ \\ \text{b.} &\text{Use the equation parameters} & a=6, h=3, k=4 \\ & \text{The vertex is the point } (h, k) &   \text{The vertex is the point } (3,4)\\ \end{array} \)

For the following quadratic functions find a. the axis of symmetry and b. the vertex

Minimum or Maximum Value of a Quadratic Function

Example \(\PageIndex{4}\)

Find the minimum or maximum value of the quadratic function \(f(x)=x^{2}+2 x-8\).

\( \begin{array}{llc}  \text{Identify the equation parameters} & a=1, b=2, c=-8 \\  \text{Since }a \text{ is positive, the parabola opens upward. }  & \text{The quadratic equation has a minimum.}  \\  \text{State the formula for the axis of symmetry } & x=-\dfrac{b}{2 a} \\  \text{Substitute and simplify.} & x=-\dfrac{2}{2 \cdot 1}= -1 \quad \text{The axis of symmetry is the line } x=-1\\ & \qquad \qquad \qquad \qquad \text{ The } x \text{ coordinate of the vertex is } x=-1\\  \text{The }y \text{ coordinate will be }f(-1) & f(-1)=({\color{red}{-1}})^2-2({\color{red}{-1}})-8 \\  \text{Simplify} & f(-1) = 1-2-8 \\  \text{The result is the }y\text{ coordinate of the vertex.} & f(-1)=-9 \qquad \text{The vertex is } (-1, -9)\\ \end{array} \)

Since the parabola has a minimum, the \(y\)-coordinate of the vertex is the minimum \(y\)-value of the quadratic equation. The minimum value of the quadratic is \(-9\) and it occurs when \(x=-1\).

Find the maximum or minimum value of the quadratic function

  • \(f(x)=x^{2}-8 x+12\).
  • \(f(x)=-4(x-2)^{2}+5\).

a. The minimum value of the quadratic function is \(−4\) and it occurs when \(x=4\).

b. The maximum value of the quadratic function is \(5\) and it occurs when \(x=2\).

Domain and Range

Any number can be the input value, \(x\), to a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all \(y\)-values greater than or equal to the \(y\)-coordinate at the vertex or less than or equal to the \(y\)-coordinate at the vertex, depending on whether the parabola opens up or down.

  • The  domain of any quadratic function is always  \( \mathbb{R}\) or \( (-\infty, \infty) \).
  • If the function is in the form \(f(x)=a(x−h)^2+k\), then the value of \(k\) is readily visible as one of the parameters.
  • If the function is in the form \(f(x)=ax^2+bx+c\), the vertex must be determined and the value for \(k\) is the \(y\) coordinate of the vertex.
  • If \(a\) is positive, the parabola has a minimum value of \(k\) and the range of the function is \( [k,\infty) \).
  • If \(a\) is negative, the parabola has a maximum value of \(k\) and the range of the function is \((-\infty, k] \).

Example \(\PageIndex{5}\): Find the Domain and Range of a Quadratic Function

Find the domain and range of \(f(x)=−5x^2+9x−1\).

As with any quadratic function, the domain is all real numbers.

Because \(a\) is negative, the parabola opens downward and has a maximum value. The maximum value must be determined. Begin by finding the \(x\)-value of the vertex.

\( h=−\dfrac{b}{2a} =−\dfrac{9}{2(-5)}=\dfrac{9}{10} \)

The maximum value is given by \(f(h)\).

\(f(\frac{9}{10})=−5(\frac{9}{10})^2+9(\frac{9}{10})-1 = \frac{61}{20}\)

The range is \(f(x){\leq}\frac{61}{20}\), or \(\left(−\infty,\frac{61}{20}\right]\).

Find the domain and range of \(f(x)=2\Big(x−\frac{4}{7}\Big)^2+\frac{8}{11}\).

The domain is all real numbers. The range is \(f(x){\geq}\frac{8}{11}\), or \(\left[\frac{8}{11},\infty\right)\).

The \(y\)-intercept is the point where the graph crosses the \(y\) axis. All points on the \(y\)-axis have an \(x\) coordinate of zero, so the \(y\)-intercept of a quadratic is found by evaluating the function \(f(0)\).

The \(x\)-intercepts are the points where the graph crosses the \(x\)-axis.  All points on the \(x\)-axis have a \(y\) coordinate of zero, so the \(x\)-intercept of a quadratic can be found by solving the equation \(f(x)=0\).  Notice in Figure \(\PageIndex{6}\) that the number of \(x\)-intercepts can vary depending upon the location of the graph.

<div data-mt-source="1"><p style="text-align: center;"><img class="internal" alt="" data-cke-saved-src="http://mathwiki.ucdavis.edu/@api/deki/files/1183/CNX_Precalc_Figure_03_02_013.jpg" src="http://mathwiki.ucdavis.edu/@api/deki/files/1183/CNX_Precalc_Figure_03_02_013.jpg" height="211" width="649"><br></p></div>

  • Evaluate \(f(0)\) to find the \(y\)-intercept. The \(y\) intercept is written in the form of a coordinate point \(0, f(0)\).
  • Solve the quadratic equation \(f(x)=0\) to find the \(x\)-intercepts. Each real number solution \(x_i\) is written as an \(x\)-intercept in the form \(x_i, 0\).

Example \(\PageIndex{6}\): Finding the \(y\)- and \(x\)-intercepts of a General Form Quadratic

Find the \(y\)- and \(x\)-intercepts of the quadratic \(f(x)=3x^2+5x−2\).

Find the \(y\)-intercept by evaluating \(f(0)\).

\( f(0)=3(0)^2+5(0)−2 =−2 \)

So the \(y\)-intercept is at \((0,−2)\).

For the \(x\)-intercepts, find all solutions of \(f(x)=0\).

\(0=3x^2+5x−2\)

In this case, the quadratic can be factored easily, providing the simplest method for solution. Typically, quadratics in general form, like this example, are usually solved using factoring, or failing that, using the quadratic formula or complete the square.

\(0=(3x−1)(x+2)\)

\[\begin{align*} 0&=3x−1 & 0&=x+2 \\ x&= \frac{1}{3} &\text{or} \;\;\;\;\;\;\;\; x&=−2 \end{align*}  \]

So the \(x\)-intercepts are at \((\frac{1}{3},0)\) and \((−2,0)\).

Example \(\PageIndex{7}\): 

Find the \(y\)- and \(x\)-intercepts of the quadratic \( f(x)=x^2+x+2\).

\( f(0)=(0)^2+(0)+2 =2 \)

So the \(y\)-intercept is at \((0, \; 2)\).

For the \(x\)-intercepts, find all solutions of \(f(x)=0\) or  \( x^2+x+2 = 0\). Clearly this does not factor, so employ the quadratic formula. 

The quadratic formula: \(x=\dfrac{−b{\pm}\sqrt{b^2−4ac}}{2a}\) and for this equation, \(a=1\), \(b=1\), and \(c=2\). Substituting these values into the formula produces

\[\begin{align*}  x&=\dfrac{−1{\pm}\sqrt{1^2−4⋅1⋅(2)}}{2⋅1} \\ &=\dfrac{−1{\pm}\sqrt{1−8}}{2} = \dfrac{−1{\pm}\sqrt{−7}}{2} =\dfrac{−1{\pm}i\sqrt{7}}{2} \nonumber \end{align*}\]

Since the solutions are imaginary, there are no \(x\)-intercepts.

Example \(\PageIndex{8}\): Find the \(y\)- and \(x\)-intercepts of a Standard Form Quadratic

Find the \(y\)- and \(x\)-intercepts of the quadratic \(f(x)=-2(x+3)^2+5\).

Find the \(y\)-intercept by evaluating \(f(0)\). Notice that the quantity inside the parentheses (\((0+3)=(3)\)) is evaluated FIRST!!!

\( \begin{align*} f(0) &=-2(0+3)^2+5\\  &=-2(3)^2+5\\  &=-2(9)+5\\  &=-18+5 = -13\\ \end{align*} \)

So the \(y\)-intercept is at \((0,−13)\).

For the \(x\)-intercepts, find all solutions of \(f(x)=0\). Solving a quadratic equation given in standard form, like in this example, is most efficiently accomplished by using the Square Root Property

\( \begin{array}{c} 0 =-2(x+3)^2+5\\ 2(x+3)^2 =5\\ (x+3)^2 =\dfrac{5}{2}\\ x+3 =\pm \sqrt{ \dfrac{5}{2}}\\ x = -3 \pm \sqrt{ 2.5 }\\ \end{array} \)

So the \(x\)-intercepts are at \( (-3+\sqrt{2.5}, 0) \) and \( (-3-\sqrt{2.5}, 0) \).

Find the \(y\)- and \(x\)-intercepts for the function \(g(x)=13+x^2−6x\).

\(y\)-intercept at \((0, 13)\), No \(x\)-intercepts

Graph a Quadratic Function

Details on how to find features of a quadratic function have been covered. Now, these features will be used to sketch a graph.

  • Determine whether the parabola opens upward \( (a > 0) \) or downward \( (a < 0) \).
  • Find the equation of the axis of symmetry, \( x = h \) where \(h=–\dfrac{b}{2a} \). 
  • Find the vertex, \( (h, k) \), where \( k = f(h) \).
  • Find the \(y\)-intercept, \( f(0) \). Find the point symmetric to the \(y\)-intercept across the axis of symmetry.
  • Find the \(x\)-intercepts. (Set \(f(x)=0\) and solve for \(x\) using factoring, QF or CTS). Find additional points if needed.
  • Graph the parabola.

Example \(\PageIndex{9}\) How to Graph a General Form Quadratic Function Using Properties

Graph \(f(x)=x^{2}-6x+8\) by using its properties.

Graph the following quadratic functions by using its properties.

  • Determine whether the parabola opens upward \( (a > 0) \) or downward \( (a < 0) \).
  • Find the equation of the axis of symmetry, \( x = h \). 
  • Find the vertex, \( (h, k) \).
  • Find the \(y\)-intercept, \( f(0) \). Find the point symmetric to the \(y\)-intercept across the axis of symmetry.
  • Find the \(x\)-intercepts. (Use the square root property to solve \(a(x-h)^2+h=0\). Find additional points if needed.

Example \(\PageIndex{10}\): How to Graph a Vertex Form Quadratic Using Properties

Graph the function \(f(x)=2(x+1)^{2}+3\) by using its properties

Graph the following functions using properties

Rewrite Quadratics into Vertex Form

As the above examples illustrate, it is often easier to graph a quadratic equation that is in vertex form, rather than in general form. This is particularly true when trying to find \(x\)-intercepts for equations that don't easily factor. There are two different approaches for transforming an equation in general form into an equation in standard (or vertex) form. One method uses the formulas for \(h\) and \(k\). The other method uses Complete the Square. Both will be illustrated below.

Formula method

  • Identify constants \( a\) and \(b\).
  • Substitute \( a\) and \(b\) into the formula: \(h=−\dfrac{b}{2a}\).
  • Substitute \(x=h\) into the general form of the quadratic function to find \(k\).
  • Rewrite the quadratic in vertex (standard) form using \(h\) and \(k\). The vertex form of the function is \(f(x)=a(x−h)^2+k \).

Example \(\PageIndex{11}\): Formula method of rewriting into vertex form

Rewrite the quadratic function \(f(x)=2x^2+4x−4\) into vertex form.

Step 1. Values of the parameters in the general form are \(a=2\), \(b=4\), and \(c=-4\).

Step 2. Solve for \(h\). 

\( h=−\dfrac{b}{2a} =−\dfrac{4}{2(2)} =−1 \)

Step 3. Use the value found for \(h\) to find \(k\).

\(  k=f(h)=f(−1) =2(−1)^2+4(−1)−4 =−6 \)

Step 4. The vertex form of the function is: 

\[\begin{align*}  f(x)&=a(x−h)^2+k \\  f(x)&=2(x+1)^2−6 \end{align*}\]

Find the vertex form for the function \(g(x)=13+x^2−6x\). 

\(g(x)=(x-3)^2 +4 \)

Complete the square method

Another way of transforming \(f(x)=ax^{2}+bx+c\) into the form \(f(x)=a(x−h)^{2}+k\) is by completing the square. The latter form is known as the vertex form or standard form. This approach will also be used when circles are studied.

We must be careful to both add and subtract the number to the SAME side of the function to complete the square. We cannot add the number to both sides as we did when we completed the square with quadratic equations.

This figure shows the difference when completing the square with a quadratic equation and a quadratic function. For the quadratic equation, start with x squared plus 8 times x plus 6 equals zero. Subtract 6 from both sides to get x squared plus 8 times x equals negative 6 while leaving space to complete the square. Then, complete the square by adding 16 to both sides to get x squared plush 8 times x plush 16 equals negative 6 plush 16. Factor to get the quantity x plus 4 squared equals 10. For the quadratic function, start with f of x equals x squared plus 8 times x plus 6. The second line shows to leave space between the 8 times x and the 6 in order to complete the square. Complete the square by adding 16 and subtracting 16 on the same side to get f of x equals x squared plus 8 times x plush 16 plus 6 minus 16. Factor to get f of x equals the quantity of x plush 4 squared minus 10.

When we complete the square in a function with a coefficient of \(x^{2}\) that is not one, we have to factor that coefficient from the \(x\)-terms. We do not factor it from the constant term. It is often helpful to move the constant term a bit to the right to make it easier to focus only on the \(x\)-terms.

Once we get the constant we want to complete the square, we must remember to multiply it by the coefficient that was part of the \(x^2\) term before we then subtract it.

  • Separate the \(x\) terms from the constant.
  • If the coefficient of \(x^{2}\) is not 1, factor it out from the \(x^2\) and \(x\) terms.
  • Find the CTS constant needed to complete the square on the  \(x^2\) and \(x\) terms. 
  • Add the CTS constant to the \(x^2\) and \(x\) terms. Subtract the CTS constant (multiplied by the coefficient of \(x^{2}\) if not 1)
  • Write the trinomial as a binomial square and combine constants outside the binomial square to arrive at the standard form of the function.

Example \(\PageIndex{12}\): CTS method of rewriting into vertex form

Rewrite \(f(x)=−3x^{2}−6x−1\) in the \(f(x)=a(x−h)^{2}+k\) form by completing the square.

Rewrite the following functions in the \(f(x)=a(x−h)^{2}+k\) form by completing the square.

Obtain the Equation of a Quadratic Function from a Graph

So far we have started with a function and then found its graph.

Now we are going to reverse the process. Starting with the graph, we will find the function.

Given a graph of a quadratic function, write the equation of the function in general form.

  • Identify the horizontal shift of the parabola; this value is \(h\). Identify the vertical shift of the parabola; this value is \(k\).
  • Substitute the values of the horizontal and vertical shift for \(h\) and \(k\). in the function \(f(x)=a(x–h)^2+k\).
  • Substitute the values \(x\) and \(f(x)\) of any point on the graph other than the vertex into the equation from step 2.
  • Solve for the parameter \(a\). The result is the quadratic equation in standard (vertex) form.
  • Expand and simplify to obtain the general form for the quadratic.

Example \(\PageIndex{13}\): Writing the Equation of a Quadratic Function from the Graph

Write an equation for the quadratic function \(g\) in Figure \(\PageIndex{13}\) in standard (vertex) form and then rewrite the result into general form.

CNX_Precalc_Figure_03_02_007.jpg

Since it is quadratic, we start with the \(g(x)=a(x−h)^{2}+k\) form. (Observe the minus sign in front of \(h\)!) The vertex, \((h,k)\), is \((−2,−3)\) so \(h=−2\) and \(k=−3\). Substituting theses values we obtain  \(g(x)=a(x+2)^2–3\).

Substituting the coordinates of a point on the curve, such as \((0,−1)\), we can solve for the parameter \(a\).

\[\begin{align*} −1&=a(0+2)^2−3 \\ 2&=4a \\ a&=\dfrac{1}{2} \end{align*}\]

Using the values found for parameters \(a,\) \(h,\) and \(k,\) write the standard form of the equation:  \(g(x)=\dfrac{1}{2}(x+2)^2–3\).

To write this in general polynomial form, we can expand the formula and simplify terms.

\[\begin{align*} g(x)&=\dfrac{1}{2}(x+2)^2−3 \\ &=\dfrac{1}{2}(x+2)(x+2)−3 \\ &=\dfrac{1}{2}(x^2+4x+4)−3 \\ &=\dfrac{1}{2}x^2+2x+2−3 \\ g(x) &=\dfrac{1}{2}x^2+2x−1 \end{align*}\]

Example \(\PageIndex{14}\)

Determine the quadratic function whose graph is shown.

The graph shown is an upward facing parabola with vertex (negative 2, negative 1) and y-intercept (0, 7).

Since it is quadratic, we start with the \(f(x)=a(x−h)^{2}+k\) form.

The vertex, \((h,k)\), is \((−2,−1)\) so \(h=−2\) and \(k=−1\).

\(f(x)=a(x-(-2))^{2}-1 \qquad \longrightarrow \qquad f(x)=a(x+2)^{2}-1\) 

To find \(a\), we use the \(y\)-intercept, \((0,7)\). So \(x=0\) and \(f(0)=7\).

\(7=a(0+2)^{2}-1\)

Solve for \(a\).

\(\begin{array}{l}{7=4 a-1} \\ {8=4 a} \\ {2=a}\end{array}\)

Write the function.

\(f(x)=2(x+2)^{2}-1\)

Write the quadratic function in the form \(f(x)=a(x−h)^{2}+k\) for each graph.

A coordinate grid has been superimposed over the quadratic path of a basketball in Figure \(\PageIndex{15}\). Find an equation for the path of the ball. Does the shooter make the basket?

CNX_Precalc_Figure_03_02_008.jpg

Figure \(\PageIndex{15}\): Stop motioned picture of a boy throwing a basketball into a hoop to show the parabolic curve it makes. (credit: modification of work by Dan Meyer)

The path passes through the origin and has vertex at \((−4, 7)\), so \(h(x)=–\frac{7}{16}(x+4)^2+7\). To make the shot, \(h(−7.5)\) would need to be about 4 but \(h(–7.5){\approx}1.64\); so, he doesn’t make it.

Key Equations

  • general form of a quadratic function: \(f(x)=ax^2+bx+c\)
  • the quadratic formula: \(x=\dfrac{−b{\pm}\sqrt{b^2−4ac}}{2a}\)
  • standard form of a quadratic function: \(f(x)=a(x−h)^2+k\)

Key Concepts

  • A polynomial function of degree two is called a quadratic function.
  • The graph of a quadratic function is a parabola. A parabola is a U-shaped curve that can open either up or down.
  • The axis of symmetry is the vertical line passing through the vertex. The \(x\)-intercepts are the points at which the parabola crosses the \(x\)-axis. The \(y\)-intercept is the point at which the parabola crosses the \(y\)-axis.
  • Quadratic functions are often written in general form. Standard or vertex form is useful to easily identify the vertex of a parabola. Either form can be written from a graph.
  • The vertex can be found from an equation representing a quadratic function. .
  • The domain of a quadratic function is all real numbers. The range varies with the function.
  • A quadratic function’s minimum or maximum value is given by the \(y\)-value of the vertex.
  • Some quadratic equations must be solved by using the quadratic formula.

axis of symmetry a vertical line drawn through the vertex of a parabola around which the parabola is symmetric; it is defined by \(x=−\dfrac{b}{2a}\).

general form of a quadratic function the function that describes a parabola, written in the form \(f(x)=ax^2+bx+c\), where \(a,b,\) and \(c\) are real numbers and a≠0.

standard form of a quadratic function the function that describes a parabola, written in the form \(f(x)=a(x−h)^2+k\), where \((h, k)\) is the vertex.

vertex the point at which a parabola changes direction, corresponding to the minimum or maximum value of the quadratic function

vertex form of a quadratic function another name for the standard form of a quadratic function

zeros in a given function, the values of \(x\) at which \(y=0\), also called roots

IMAGES

  1. Graphing Quadratic Functions In Standard Form Worksheet

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  2. Using The Quadratic Formula Worksheet

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  3. Unit 8 Quadratic Equations Homework 2 Intro To Quadratics

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  4. Intro To Quadratics Worksheet

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  5. Quadratic equations

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  6. Identifying Quadratic Equations Worksheet Pdf

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VIDEO

  1. The length of a rectangle is 3 times it’s width, if the area is 48 what is the L and W?

  2. Graphing Quadratic Equation using Table of Values (tagalog)

  3. Algebra 2 Lesson: Solving Quadratic Equations by Graphing

  4. Graphing Quadratic Equations: Vertical Reflection and Stretch

  5. A19.19 Solving Quadratic Equations by Graphing

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  3. Free Printable Math Worksheets for Algebra 2

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    The graph of a quadratic function is a parabola, which is a "u"-shaped curve: In this article, we review how to graph quadratic functions. Looking for an introduction to parabolas? Check out this video. Example 1: Vertex form Graph the equation. y = − 2 ( x + 5) 2 + 4 This equation is in vertex form. y = a ( x − h) 2 + k

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  9. 10.5 Graphing Quadratic Equations in Two Variables

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  10. Unit 8 Homework 2 Graphing Quadratic Equations

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  23. 3.1: Graphs of Quadratic Functions

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