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10.5: Graphing Quadratic Equations
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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
Before you get started, take this readiness quiz.
- Graph the equation \(y=3x−5\) by plotting points. If you missed this problem, review [link] .
- Evaluate \(2x^2+4x−1\)when \(x=−3\) If you missed this problem, review [link] .
- Evaluate \(−\frac{b}{2a}\) when \(a=13\) and b=\(\frac{5}{6}\) If you missed this problem, review [link] .
Recognize the Graph of a Quadratic Equation in Two Variables
We have graphed equations of the form \(Ax+By=C\). We called equations like this linear equations because their graphs are straight lines.
Now, we will graph equations of the form \(y=ax^2+bx+c\). We call this kind of equation a quadratic equation in two variables .
definition: QUADRATIC EQUATION IN TWO VARIABLES
A quadratic equation in two variables , where a,b,and c are real numbers and \(a\neq 0\), is an equation of the form \[y=ax^2+bx+c \nonumber\]
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 \(y=x^2\). We will choose integer values of x between −2 and 2 and find their y values. See Table .
Notice when we let \(x=1\) and \(x=−1\), we got the same value for y.
\[\begin{array} {ll} {y=x^2} &{y=x^2} \\ {y=1^2} &{y=(−1)^2} \\ {y=1} &{y=1} \\ \nonumber \end{array}\]
The same thing happened when we let \(x=2\) and \(x=−2\).
Now, we will plot the points to show the graph of \(y=x^2\). See Figure .
The graph is not a line. This figure is called a parabola . Every quadratic equation has a graph that looks like this.
In Example you will practice graphing a parabola by plotting a few points.
Example \(\PageIndex{1}\)
\(y=x^2-1\)
We will graph the equation by plotting points.
Example \(\PageIndex{2}\)
Graph \(y=−x^2\).
Example \(\PageIndex{3}\)
Graph \(y=x^2+1\).
How do the equations \(y=x^2\) and \(y=x^2−1\) differ? What is the difference between their graphs? How are their graphs the same?
All parabolas of the form \(y=ax^2+bx+c\) open upwards or downwards. See Figure .
Notice that the only difference in the two equations is the negative sign before the \(x^2\) in the equation of the second graph in Figure . When the \(x^2\) term is positive, the parabola opens upward, and when the \(x^2\) term is negative, the parabola opens downward.
Definition: PARABOLA ORIENTATION
For the quadratic equation \(y=ax^2+bx+c\), if:
Example\(\PageIndex{4}\)
Determine whether each parabola opens upward or downward:
- \(y=−3x^2+2x−4\)
- \( y=6x^2+7x−9\)
Example\(\PageIndex{5}\)
- \(y=2x^2+5x−2\)
- \(y=−3x^2−4x+7\)
Example \(\PageIndex{6}\)
- \(y=−2x^2−2x−3\)
- \(y=5x^2−2x−1\)
Find the Axis of Symmetry and Vertex of a Parabola
Look again at Figure . 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 red. See Figure .
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 \(y=ax^2+bx+c\) is x=\(−\frac{b}{2a}\).
So, to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula x=\(−\frac{b}{2a}\).
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 \(−\frac{b}{2a}\). To find the y -coordinate of the vertex, we substitute the value of the x -coordinate into the quadratic equation.
Definition: AXIS OF SYMMETRY AND VERTEX OF A PARABOLA
For a parabola with equation \(y=ax^2+bx+c\):
- The axis of symmetry of a parabola is the line x=\(−\frac{b}{2a}\).
- The vertex is on the axis of symmetry, so its x -coordinate is \(−\frac{b}{2a}\).
To find the y -coordinate of the vertex, we substitute x=\(−\frac{b}{2a}\) into the quadratic equation.
Example\(\PageIndex{7}\)
For the parabola \(y=3x^2−6x+2\) find:
- the axis of symmetry and
- the vertex.
Example \(\PageIndex{8}\)
For the parabola \(y=2x^2−8x+1\) find:
- (2,−7)
Example \(\PageIndex{9}\)
For the parabola \(y=2x^2−4x−3\) find:
- (1,−5)
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 is zero. So, to find the y -intercept, we substitute 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 is zero. To find an x -intercept, we substitute \(y=0\) into the equation. In other words, we will need to solve the equation \(0=ax^2+bx+c\) for x.
\[\begin{array} {ll} {y=ax^2+bx+c} \\ {0=ax^2+bx+c} \\ \nonumber \end{array}\]
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 .
First, we will find the x -intercepts of a parabola with equation \(y=x^2+4x+3\).
Now, we will find the x -intercepts of the parabola with equation \(y=−x^2+4x+3\).
We will use the decimal approximations of the x-intercepts, so that we can locate these points on the graph.
\[\begin{array} {l} {(2+\sqrt{7},0) \approx (4.6,0)} & {(2−\sqrt{7},0) \approx (-0.6,0)}\\ \nonumber \end{array}\]
Do these results agree with our graphs? See Figure .
Definition: FIND THE INTERCEPTS OF A PARABOLA
To find the intercepts of a parabola with equation \(y=ax^2+bx+c\):
\[\begin{array}{ll} {\textbf{y-intercept}}& {\textbf{x-intercept}}\\ {\text{Let} x=0 \text{and solve the y}}& {\text{Let} y=0 \text{and solve the x}}\\ \nonumber \end{array}\]
Example \(\PageIndex{10}\)
Find the intercepts of the parabola \(y=x^2−2x−8\).
Example \(\PageIndex{11}\)
Find the intercepts of the parabola \(y=x^2+2x−8\).
y:(0,−8); x:(−4,0), (2,0)
Example \(\PageIndex{12}\)
Find the intercepts of the parabola \(y=x^2−4x−12\).
y:(0,−12); x:(6,0), (−2,0)
In this chapter, we have been solving quadratic equations of the form \(ax^2+bx+c=0\). We solved for xx and the results were the solutions to the equation.
We are now looking at quadratic equations in two variables of the form \(y=ax^2+bx+c\). The graphs of these equations are parabolas. The x -intercepts of the parabolas occur where y=0.
For example:
\[\begin{array}{cc} {\textbf{Quadratic equation}}&{\textbf{Quadratic equation in two variable}}\\ {}&{y=x^2−2x−15}\\ {x^2−2x−15}&{\text{Let} y=0, 0=x^2−2x−15}\\ {(x−5)(x+3)=0}&{0=(x−5)(x+3)}\\ {x−5=0, x+3=0}&{x−5=0, x+3=0}\\ {x=5, x=−3}&{x=5, x=−3}\\ {}&{(5,0) \text{and} (−3,0)}\\ {}&{\text{x-intercepts}}\\ \end{array}\]
The solutions of the quadratic equation are the 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 \(ax^2+bx+c=0\). 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 \(\PageIndex{13}\)
Find the intercepts of the parabola \(y=5x^2+x+4\).
Example \(\PageIndex{14}\)
Find the intercepts of the parabola \(y=3x^2+4x+4\).
y:(0,4); x:none
Example \(\PageIndex{15}\)
Find the intercepts of the parabola \(y=x^2−4x−5\).
y:(0,−5); x:(5,0)(−1,0)
Example \(\PageIndex{16}\)
Find the intercepts of the parabola \(y=4x^2−12x+9\).
Example \(\PageIndex{17}\)
Find the intercepts of the parabola \(y=−x^2−12x−36.\).
y:(0,−36); x:(−6,0)
Example \(\PageIndex{18}\)
Find the intercepts of the parabola \(y=9x^2+12x+4\).
y:(0,4); x:\((−\frac{2}{3},0)\)
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.
How To Graph a Quadratic Equation in Two Variables
Example \(\PageIndex{19}\)
Graph \(y=x2−6x+8\).
Example \(\PageIndex{20}\)
Graph the parabola \(y=x^2+2x−8\).
y:(0,−8); x:(2,0),(−4,0); axis: x=−1; vertex: (−1,−9);
Example \(\PageIndex{21}\)
Graph the parabola \(y=x^2−8x+12\).
y:(0,12); x:(2,0),(6,0); axis: x=4; vertex:(4,−4);
Definition: GRAPH A QUADRATIC EQUATION IN TWO VARIABLES.
- Write the quadratic equation with yy on one side.
- Determine whether the parabola opens upward or downward.
- Find the axis of symmetry.
- Find the vertex.
- Find the y -intercept. Find the point symmetric to the y -intercept across the axis of symmetry.
- Find the x -intercepts.
- 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 \(\PageIndex{22}\)
Graph \(y=−x^2+6x−9\).
Example \(\PageIndex{23}\)
Graph the parabola \(y=−3x^2+12x−12\).
y:(0,−12); x:(2,0); axis: x=2; vertex:(2,0);
Example \(\PageIndex{24}\)
Graph the parabola \(y=25x^2+10x+1\).
y:(0,1); x:(−15,0); axis: x=−15; vertex:(−15,0);
For the graph of \(y=−x^2+6x−9\) 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 \(0=−x^2+6x−9\) 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 \(y=x^2+4x+5\)?
Example \(\PageIndex{25}\)
Graph \(y=x^2+4x+5\).
Example \(\PageIndex{26}\)
Graph the parabola \(y=2x^2−6x+5\).
y:(0,5); x:none; axis: \(x=\frac{3}{2}\); vertex:\((\frac{3}{2},\frac{1}{2})\);
Example \(\PageIndex{27}\)
Graph the parabola \(y=−2x^2−1\).
y:(0,−1); x:none; axis: x=0; vertex:(0,−1);
Finding the y -intercept by substituting 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 . We will use the Quadratic Formula again in the next example.
Example \(\PageIndex{28}\)
Graph \(y=2x^2−4x−3\).
Example \(\PageIndex{29}\)
Graph the parabola \(y=5x^2+10x+3\).
y:(0,3); x:(−1.6,0),(−0.4,0); axis: x=−1; vertex:(−1,−2);
Example \(\PageIndex{30}\)
Graph the parabola \(y=−3x^2−6x+5\).
y:(0,5); x:(0.6,0),(−2.6,0); axis: x=−1; vertex:(−1,8);
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 .
Definition: 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 \(\PageIndex{31}\)
Find the minimum value of the quadratic equation \(y=x^2+2x−8\).
Example \(\PageIndex{32}\)
Find the maximum or minimum value of the quadratic equation \(y=x^2−8x+12\).
The minimum value is −4 when x=4.
Example \(\PageIndex{33}\)
Find the maximum or minimum value of the quadratic equation \(y=−4x^2+16x−11\).
The maximum value is 5 when x=2.
We have used the formula
\[\begin{array} {l} {h=−16t^2+v_{0}t+h_{0}}\\ \nonumber \end{array}\]
to calculate the height in feet, h, of an object shot upwards into the air with initial velocity, \(v_{0}\), after t seconds.
This formula is a quadratic equation in the variable tt, 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 \(\PageIndex{34}\)
The quadratic equation \(h=−16t^2+v_{0}t+h_{0}\) 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.
\(h=−16t^2+176t+4\)
Since a is negative, the parabola opens downward.
The quadratic equation has a maximum.
1. \[\begin{array} {ll} {}&{t=−\frac{b}{2a}}\\ {\text{Find the axis of symmetry.}}& {t=−\frac{176}{2(−16)}}\\ {}&{t=5.5}\\ {}&{\text{The axis of symmetry is} t = 5.5}\\ {\text{The vertex is on the line} t=5.5}& {\text{The maximum occurs when} t =5.5 \text{seconds.}}\\ \nonumber \end{array}\]
Example \(\PageIndex{35}\)
The quadratic equation \(h=−16t^2+128t+32\) 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.
It will take 4 seconds to reach the maximum height of 288 feet.
Example \(\PageIndex{36}\)
A toy rocket shot upward from the ground at a rate of 208 ft/sec has the quadratic equation of \(h=−16t^2+208t\). When will the rocket reach its maximum height? What will be the maximum height? Round answers to the nearest tenth.
It will take 6.5 seconds to reach the maximum height of 676 feet.
- Graphing Quadratic Functions
- How do you graph a quadratic function?
- Graphing Quadratic Equations
Key Concepts
- The graph of every quadratic equation is a parabola.
- a>0, the parabola opens upward.
- a<0, the parabola opens downward.
- The axis of symmetry of a parabola is the line \(x=−\frac{b}{2a}\).
- To find the y -coordinate of the vertex we substitute \(x=−\frac{b}{2a}\) into the quadratic equation.
- Find the Intercepts of a Parabola To find the intercepts of a parabola with equation \(y=ax^2+bx+c\): \[\begin{array} {ll} {\textbf{y-intercept}}&{\textbf{x-intercepts}}\\ {\text{Let} x=0 \text{and solve for y}}&{\text{Let} y=0 \text{and solve for x}}\\ \nonumber \end{array}\]
- The y - coordinate of the vertex of the graph of a quadratic equation is the
Graphing Quadratics in Vertex Form Algebra 1 Practice Worksheet
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This 8-question algebra 1 worksheet provides students with organized practice graphing quadratics in Vertex Form. Students will first identify the vertex and then complete a table of values in order to graph the parabola. I have also provided an option that does not include the function table.
Works great as a class work or homework assignment after using this ★FOLDABLE★ to introduce the concept.
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Finding angles of triangles. Finding side lengths of triangles. Statistics. Visualizing data. Center and spread of data. Scatter plots. Using statistical models. Free Algebra 1 worksheets created with Infinite Algebra 1. Printable in convenient PDF format.
Check out this video. Example: Non-vertex form Graph the function. g ( x) = x 2 − x − 6 First, let's find the zeros of the function—that is, let's figure out where this graph y = g ( x) intersects the x -axis. g ( x) = x 2 − x − 6 0 = x 2 − x − 6 0 = ( x − 3) ( x + 2)
Familiar Attempted Not started Quiz Unit test About this unit We've seen linear and exponential functions, and now we're ready for quadratic functions. We'll explore how these functions and the parabolas they produce can be used to solve real-world problems. Intro to parabolas Learn Parabolas intro Interpreting a parabola in context
Algebra (all content) 20 units · 412 skills. Unit 1 Introduction to algebra. Unit 2 Solving basic equations & inequalities (one variable, linear) Unit 3 Linear equations, functions, & graphs. Unit 4 Sequences. Unit 5 System of equations. Unit 6 Two-variable inequalities. Unit 7 Functions. Unit 8 Absolute value equations, functions, & inequalities.
Find the vertex. Is the vertex a maximum or a minimum? Can you tell (without graphing) if your vertex is going to be a maximum or a minimum? a) y = x2 b) y =1 2 x2c) y = ! 2 x2 1 Example 3: Without graphing, order the quadratic functions from widest to narrowest: y = ! 4 x2,y= 1 4 x2,y= x2 Example 4: Graph the following functions.
Infinite Algebra 1 covers all typical algebra material, over 90 topics in all, from adding and subtracting positives and negatives to solving rational equations. Suitable for any class with algebra content. Designed for all levels of learners from remedial to advanced. Test and worksheet generator for Algebra 1.
Answer. We will graph the equation by plotting points. Choose integers values for x, substitute them into the equation and solve for y. Plot the points, and then connect them with a smooth curve. The result will be the graph of the equation \ (y=x^2−1\) Example \ (\PageIndex {2}\) Graph \ (y=−x^2\). Answer.
This Algebra 1 - Quadratic Functions Worksheets will produce problems for practicing graphing quadratic function from their equations. These Quadratic Functions Worksheets are a good resource for students in the 5th Grade through the 8th Grade. Graphing Quadratic Inequalities Worksheets
L 3 GMca sd de4 Ew 0i vt ihe WIDnlf Lijn uiJtIe R yA8l Ggke 8brsa V f13. b Worksheet by Kuta Software LLC Algebra 1 ID: 1 ... Graphing Quadratics Extra Practice List the concavity, y-intercept, axis of symmetry, and vertex for each function. 1) y = x2 + 8x + 15 2) y = −3x2 − 24 x − 49
Find a quadratic model for each set of values. 1.1. (-1, 1), (1, 1), (3, 9) 2. (-4, 8), (-1, 5), (1, 13) 3. (-1, 10), (2, 4), (3,-6) 4.
8. Solutions to quadratic equations are called _______________________. Determine whether the quadratic functions have two real roots, one real root, or no real roots. If possible, list the zeros of the function. 9. Number of roots: _____ 10. Number of roots: _____ 11. Number of roots: ____ Zero(s): ____________ Zero(s): ____________
Lesson 10: Quadratic standard form. Finding the vertex of a parabola in standard form. Graphing quadratics: standard form. Math >. Algebra 1 >. Quadratic functions & equations >.
Common Core Standard: A-SSE.A.2, A-SSE.B.3, A-APR.B.3, F-IF.B.4, F-IF.C.7, F-IF.C.8
This 8-question algebra 1 worksheet provides students with organized practice graphing quadratics in Vertex Form. Students will first identify the vertex and then complete a table of values in order to graph the parabola. I have also provided an option that does not include the function table. Works great as a class work or homework assignment ...
Download Free Algebra I > Graphing Quadratics Worksheets Below: All worksheets are free to download and use for practice or in your classroom. All we ask is that you don't remove the KidSmart logo. Click on for Answers Graphing Quadratics Graphing Quadratics in Vertex Form - Graphing Quadratics in Intercept Form -
Completing the Square Worksheet #1. Completing the Square Worksheet #2. Solve by Graphing Worksheet and Review - To solve by graphing, the answers come from where the curved line crosses the x-axis. All Graphs are provided.
CHAPTER 2 WORKSHEETS. F ractions Review WS # 1 (Solns on back of WS) 2-1 Solving One-Step Equation s. 2-2 Solving Two-Step Equations. 2-3 Solving Multi-Step Equations . 2- 4 Solving Equations with Variables on Both Sides ( SOLUTIONS) 2-5 Literal Equations and Formulas. 2-6 Ratios, Rates, and Conversions ( SOLUTIONS)
Enjoy these free printable sheets. Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. Plus each one comes with an answer key. Algebra. Distance Formula. Equation of Circle. Factoring. Factor Trinomials Worksheet. Functions and Relations.
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