Elementary Algebra 2e · free preview

10.5 Graphing Quadratic Equations in Two Variables

Learning objectives: 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

1. Learning objectives

Learning objectives

Learning objectives

  • 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

2. Recognize the Graph of a Quadratic Equation in Two Variables

Recognize the Graph of a Quadratic Equation in Two Variables

We have graphed equations of the form Ax+By=CAx+By=C. We called equations like this linear equations because their graphs are straight lines.

Now, we will graph equations of the form y=ax2+bx+cy=a{x}^{2}+bx+c. We call this kind of equation a quadratic equation in two variables.

A quadratic equation in two variables, where a,b, and ca,b,\,\mathrm{and}\,c are real numbers and a≠0a\ne{}0, is an equation of the form

y=ax2+bx+cy=a{x}^{2}+bx+c

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=x2y={x}^{2}. We will choose integer values of xx between −2-2 and 2 and find their yy values. See Table 10.1.

y=x2y={x}^{2}
xxyy
00
11
−1-11
24
−2-24

Table

10.1

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

y=x2y=x2y=12y=(−1)2y=1y=1\begin{array}{lccl} y={x}^{2} & & & y={x}^{2} \\ y={1}^{2} & & & y={(-1)}^{2} \\ y=1 & & & y=1 \\ \end{array}

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

Now, we will plot the points to show the graph of y=x2y={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.

3. Example 10.43

Example 10.43

Example 10.43

Graph y=x2−1y={x}^{2}-1.

Solution

We will graph the equation by plotting points.

How do the equations y=x2y={x}^{2} and y=x2−1y={x}^{2}-1 differ? What is the difference between their graphs? How are their graphs the same?

All parabolas of the form y=ax2+bx+cy=a{x}^{2}+bx+c open upwards or downwards. See Figure 10.3.

Notice that the only difference in the two equations is the negative sign before the x2{x}^{2} in the equation of the second graph in Figure 10.3. When the x2{x}^{2} term is positive, the parabola opens upward, and when the x2{x}^{2} term is negative, the parabola opens downward.

For the quadratic equation y=ax2+bx+cy=a{x}^{2}+bx+c, if:

4. Example 10.44

Example 10.44

Example 10.44

Determine whether each parabola opens upward or downward:

ⓐ y=−3x2+2x−4y=-3{x}^{2}+2x-4 ⓑ y=6x2+7x−9y=6{x}^{2}+7x-9

Solution

5. Find the Axis of Symmetry and Vertex of a Parabola

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=ax2+bx+cy=a{x}^{2}+bx+c is x=−b2a.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=−b2ax=-\frac{b}{2a}.

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 −b2a-\frac{b}{2a}. To find the y-coordinate of the vertex, we substitute the value of the x-coordinate into the quadratic equation.

For a parabola with equation y=ax2+bx+cy=a{x}^{2}+bx+c:

  • The axis of symmetry of a parabola is the line x=−b2ax=-\frac{b}{2a}.
  • The vertex is on the axis of symmetry, so its x-coordinate is −b2a-\frac{b}{2a}.

To find the y-coordinate of the vertex, we substitute x=−b2ax=-\frac{b}{2a} into the quadratic equation.

6. Example 10.45

Example 10.45

Example 10.45

For the parabola y=3x2−6x+2y=3{x}^{2}-6x+2 find: ⓐ the axis of symmetry and ⓑ the vertex.

Solution

7. Find the Intercepts of a Parabola

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 xx is zero. So, to find the y-intercept, we substitute x=0x=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 yy is zero. To find an x-intercept, we substitute y=0y=0 into the equation. In other words, we will need to solve the equation 0=ax2+bx+c0=a{x}^{2}+bx+c for xx.

y=ax2+bx+c0=ax2+bx+c\begin{array}{l} y=a{x}^{2}+bx+c \\ 0=a{x}^{2}+bx+c \\ \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 10.5.

First, we will find the x-intercepts of a parabola with equation y=x2+4x+3y={x}^{2}+4x+3.

Now, we will find the x-intercepts of the parabola with equation y=−x2+4x+3y=-{x}^{2}+4x+3.

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

(2+7,0)≈(4.6,0)(2−7,0)≈(−0.6,0)\begin{array}{lccl} (2+\sqrt{7},0)\approx(4.6,0) & & & (2-\sqrt{7},0)\approx(-0.6,0) \\ \end{array}

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=ax2+bx+cy=a{x}^{2}+bx+c:

y−intercept x−interceptsLet x=0 andsolvefor y. Let y=0 andsolvefor x.\begin{array}{cccc} y\mathrm{-intercept} & & & \,x\mathrm{-intercepts} \\ \mathrm{Let}\,x=0\,\mathrm{and solve for}\,y. & & & \,\mathrm{Let}\,y=0\,\mathrm{and solve for}\,x. \\ \end{array}

8. Example 10.46

Example 10.46

Example 10.46

Find the intercepts of the parabola y=x2−2x−8y={x}^{2}-2x-8.

Solution

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

In this chapter, we have been solving quadratic equations of the form ax2+bx+c=0a{x}^{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=ax2+bx+cy=a{x}^{2}+bx+c. The graphs of these equations are parabolas. The x-intercepts of the parabolas occur where y=0y=0.

For example:

Quadraticequation Quadraticequationintwovariables y=x2−2x−15x2−2x−15=0(x−5)(x+3)=0 let y=0 0=x2−2x−150=(x−5)(x+3)x−5=0x+3=0x=5x=−3 x−5=0x+3=0x=5x=−3 (5,0) and (−3,0) x−intercepts\begin{array}{cccccccccccccccccccccccccccccccc} \mathrm{Quadratic equation} & & & \,\mathrm{Quadratic equation in two variables} \\ & & & \,y={x}^{2}-2x-15 \\ \begin{array}{rll} {x}^{2}-2x-15 & = & 0 \\ (x-5)(x+3) & = & 0 \\ \end{array} & & & \,\mathrm{let}\,y=0\,\begin{array}{l} 0={x}^{2}-2x-15 \\ 0=(x-5)(x+3) \\ \end{array} \\ \begin{array}{rllccrll} x-5 & = & 0 & & & x+3 & = & 0 \\ x & = & 5 & & & x & = & -3 \\ \end{array} & & & \,\begin{array}{rllccrll} x-5 & = & 0 & & & x+3 & = & 0 \\ x & = & 5 & & & x & = & -3 \\ \end{array} \\ & & & \,(5,0)\,\mathrm{and}\,(-3,0) \\ & & & \,x\mathrm{-intercepts} \\ \end{array}

The solutions of the quadratic equation are the xx 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 ax2+bx+c=0a{x}^{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.

9. Example 10.47

Example 10.47

Example 10.47

Find the intercepts of the parabola y=5x2+x+4y=5{x}^{2}+x+4.

Solution

10. Example 10.48

Example 10.48

Example 10.48

Find the intercepts of the parabola y=4x2−12x+9y=4{x}^{2}-12x+9.

Solution

11. Graph Quadratic Equations in Two Variables

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.

12. Example 10.49

Example 10.49

Example 10.49

How To Graph a Quadratic Equation in Two Variables

Graph y=x2−6x+8y={x}^{2}-6x+8.

Graph a quadratic equation in two variables.

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

13. Example 10.50

Example 10.50

Example 10.50

Graph y=−x2+6x−9y=-{x}^{2}+6x-9.

Solution

For the graph of y=−x2+6x−9y=-{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=−x2+6x−90=-{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=x2+4x+5y={x}^{2}+4x+5?

14. Example 10.51

Example 10.51

Example 10.51

Graph y=x2+4x+5y={x}^{2}+4x+5.

Solution

Finding the y-intercept by substituting x=0x=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.

15. Example 10.52

Example 10.52

Example 10.52

Graph y=2x2−4x−3y=2{x}^{2}-4x-3.

Solution

16. Solve Maximum and Minimum Applications

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.

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.

17. Example 10.53

Example 10.53

Example 10.53

Find the minimum value of the quadratic equation y=x2+2x−8y={x}^{2}+2x-8.

Solution

We have used the formula

h=−16t2+v0t+h0h=-16{t}^{2}+{v}_{0}t+{h}_{0}

to calculate the height in feet, hh, of an object shot upwards into the air with initial velocity, v0{v}_{0}, after tt 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.

18. Example 10.54

Example 10.54

Example 10.54

The quadratic equation h=−16t2+v0t+h0h=-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.

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

Solution

h=−16t2+176t+4h=-16{t}^{2}+176t+4

Since a is negative, the parabola opens downward.

The quadratic equation has a maximum.

  1. ⓐ Findtheaxisofsymmetry. t=−b2at=−1762(−16)t=5.5 Theaxisofsymmetryis t=5.5.Thevertexisontheline t=5.5. Themaximumoccurswhen t=5.5 seconds.\begin{array}{lccl} \mathrm{Find the axis of symmetry.} & & & \,\begin{array}{l} t=-\frac{b}{2a} \\ t=-\frac{176}{2(-16)} \\ t=5.5 \\ \end{array} \\ & & & \,\mathrm{The axis of symmetry is}\,t=5.5. \\ \mathrm{The vertex is on the line}\,t=5.5. & & & \,\mathrm{The maximum occurs when}\,t=5.5\,\mathrm{seconds.} \\ \end{array}
  1. ⓑ Find h when t=5.5t=5.5. Use a calculator to simplify. The vertex is (5.5,488)(5.5,488). Since the parabola has a maximum, the *h-*coordinate of the vertex is the maximum y-value of the quadratic equation. The maximum value of the quadratic is 488 feet and it occurs when t=5.5t=5.5 seconds.

19. Section 10.5 Exercises

Section 10.5 Exercises

Section 10.5 Exercises

Recognize the Graph of a Quadratic Equation in Two Variables

In the following exercises, graph:

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

Find the Axis of Symmetry and Vertex of a Parabola

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

Find the Intercepts of a Parabola

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

Graph Quadratic Equations in Two Variables

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

Solve Maximum and Minimum Applications

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

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

Everyday Math

Writing Exercises

ⓐ 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?

20. Practice 1

  1. Graph y=−x2y=-{x}^{2}.
  1. Graph y=x2+1y={x}^{2}+1.
  1. Determine whether each parabola opens upward or downward:

ⓐ y=2x2+5x−2y=2{x}^{2}+5x-2 ⓑ y=−3x2−4x+7y=-3{x}^{2}-4x+7

  1. Determine whether each parabola opens upward or downward:

ⓐ y=−2x2−2x−3y=-2{x}^{2}-2x-3 ⓑ y=5x2−2x−1y=5{x}^{2}-2x-1

21. Practice 2

  1. For the parabola y=2x2−8x+1y=2{x}^{2}-8x+1 find: ⓐ the axis of symmetry and ⓑ the vertex.
  1. For the parabola y=2x2−4x−3y=2{x}^{2}-4x-3 find: ⓐ the axis of symmetry and ⓑ the vertex.
  1. Find the intercepts of the parabola y=x2+2x−8.y={x}^{2}+2x-8.
  1. Find the intercepts of the parabola y=x2−4x−12.y={x}^{2}-4x-12.

22. Practice 3

  1. Find the intercepts of the parabola y=3x2+4x+4.y=3{x}^{2}+4x+4.
  1. Find the intercepts of the parabola y=x2−4x−5.y={x}^{2}-4x-5.
  1. Find the intercepts of the parabola y=−x2−12x−36.y=-{x}^{2}-12x-36.
  1. Find the intercepts of the parabola y=9x2+12x+4.y=9{x}^{2}+12x+4.

23. Practice 4

  1. Graph the parabola y=x2+2x−8.y={x}^{2}+2x-8.
  1. Graph the parabola y=x2−8x+12.y={x}^{2}-8x+12.
  1. Graph the parabola y=−3x2+12x−12.y=-3{x}^{2}+12x-12.
  1. Graph the parabola y=25x2+10x+1.y=25{x}^{2}+10x+1.

24. Practice 5

  1. Graph the parabola y=2x2−6x+5.y=2{x}^{2}-6x+5.
  1. Graph the parabola y=−2x2−1.y=-2{x}^{2}-1.
  1. Graph the parabola y=5x2+10x+3.y=5{x}^{2}+10x+3.
  1. Graph the parabola y=−3x2−6x+5.y=-3{x}^{2}-6x+5.

25. Practice 6

  1. Find the maximum or minimum value of the quadratic equation y=x2−8x+12y={x}^{2}-8x+12.
  1. Find the maximum or minimum value of the quadratic equation y=−4x2+16x−11y=-4{x}^{2}+16x-11.
  1. The quadratic equation h=−16t2+128t+32h=-16{t}^{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.
  1. A toy rocket shot upward from the ground at a rate of 208 ft/sec has the quadratic equation of h=−16t2+208th=-16{t}^{2}+208t. When will the rocket reach its maximum height? What will be the maximum height? Round answers to the nearest tenth.

26. Practice 7

  1. y=x2+3y={x}^{2}+3
  1. y=−2x2−6x−7y=-2{x}^{2}-6x-7
  1. y=4x2+x−4y=4{x}^{2}+x-4
  1. y=x2+8x−1y={x}^{2}+8x-1

27. Practice 8

  1. y=−x2+2x+5y=-{x}^{2}+2x+5
  1. y=x2+7x+6y={x}^{2}+7x+6
  1. y=−x2+8x−19y=-{x}^{2}+8x-19
  1. y=4x2−20x+25y=4{x}^{2}-20x+25

28. Practice 9

  1. y=x2+6x+5y={x}^{2}+6x+5
  1. y=x2+4x+3y={x}^{2}+4x+3
  1. y=9x2+12x+4y=9{x}^{2}+12x+4
  1. y=−x2+2x−7y=-{x}^{2}+2x-7

29. Practice 10

  1. y=2x2−4x+1y=2{x}^{2}-4x+1
  1. y=2x2−4x+2y=2{x}^{2}-4x+2
  1. y=−x2−4x+2y=-{x}^{2}-4x+2
  1. y=5x2−10x+8y=5{x}^{2}-10x+8

30. Practice 11

  1. y=3x2+18x+20y=3{x}^{2}+18x+20
  1. y=2x2+x−1y=2{x}^{2}+x-1
  1. y=x2−6x+15y={x}^{2}-6x+15
  1. y=−9x2+16y=-9{x}^{2}+16

31. Practice 12

  1. An arrow is shot vertically upward from a platform 45 feet high at a rate of 168 ft/sec. Use the quadratic equation h=−16t2+168t+45h=-16{t}^{2}+168t+45 to find how long it will take the arrow to reach its maximum height, and then find the maximum height.
  1. A computer store owner estimates that by charging xx dollars each for a certain computer, he can sell 40−x40-x computers each week. The quadratic equation R=−x2+40xR=-{x}^{2}+40x is used to find the revenue, RR, received when the selling price of a computer is xx. Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.
  1. 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(240−2x)A=x(240-2x) gives the area of the corral, AA, 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.
  1. In the previous set of exercises, you worked with the quadratic equation R=−x2+40xR=-{x}^{2}+40x that modeled the revenue received from selling computers at a price of xx 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=−x2+40xR=-{x}^{2}+40x. ⓑ Find the values of the x-intercepts.

32. Practice 13

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

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