Section 2.2 Graphs of Equations

Before Starting this Section, Review

  1. 1 How to plot points (Section 2.1 , page 176)

  2. 2 Equivalent equations (Section 1.1 , page 83)

  3. 3 Completing squares (Section 1.3 , page 109)

Objectives

  1. 1 Sketch a graph by plotting points.

  2. 2 Find the intercepts of a graph.

  3. 3 Find the symmetries in a graph.

  4. 4 Find the equation of a circle.

Leafsmen Discover Deer

A herd of 400 deer was introduced onto a small island called Leafs. The natives both liked and admired these beautiful creatures. However, the Leafsmen soon discovered that deer meat is excellent food. Also, a rumor spread throughout Leafs that eating deer meat prolonged one’s life. The natives then began to hunt the deer. The number of deer, y, after t years from the initial introduction of deer into Leafs is described by the equation y=−t4+96t2+400.y=−t4+96t2+400. When does the population of deer become extinct in Leafs? (See Example 5.)

Graph of an Equation

  1. 1 Sketch a graph by plotting points.

An equation is an algebraic equality relating one or more quantities. An equation involving two unknown quantities describes a relation between these two quantities and specifies how one quantity changes with respect to the other quantity. The two changing (or varying) quantities are often represented by variables. The following equations are examples of relationships between two variables:

y=2x+1; x2+y2=4; y=x2; x=y2; F=95 C+32;and q=−3p2+30.
y=2x+1; x2+y2=4; y=x2; x=y2; F=95 C+32;and q=−3p2+30.

An ordered pair (a, b) is said to satisfy an equation with variables x and y if, when a is substituted for x and b is substituted for y in the equation, the resulting statement is true. For example, the ordered pair (2, 5) satisfies the equation y=2x+1y=2x+1 because replacing x with 2 and y with 5 yields 5=2(2)+1,5=2(2)+1, and simplifies to 5=5,5=5, which is a true statement. The ordered pair (5,−2)(5,−2) does not satisfy this equation since replacing x with 5 and y with −2−2 yields −2=2(5)+1,−2=2(5)+1, and simplifies to −2=11−2=11, which is false. An ordered pair that satisfies an equation is called a solution of the equation.

In an equation involving x and y, if the value of y can be found given the value of x, then we say that y is the dependent variable and x is the independent variable. In the equation y=2x+1,y=2x+1, for any real number x, there is a corresponding value of y. Hence, we have infinitely many solutions of the equation y=2x+1.y=2x+1. When these solutions are graphed or plotted as points in the coordinate plane, they constitute the graph of the equation.

The graph of an equation is a geometric visualization of its solution set. In short, the coordinate place allows us to investigate algebraic equations geometrically and to read many properties directly from their graphs.

Example 1 Sketching a Graph by Plotting Points

Sketch the graph of y=x2−3.y=x2−3.

Solution

There are infinitely many solutions of the equation y=x2−3.y=x2−3. To find a few, we choose integer values of x between −3−3 and 3. Then we find the corresponding values of y as shown in Table 2.3.

Table 2.3

x y=x2−3y=x2−3 (x, y)
−3−3 y=(−3)2−3=9−3=6y=(−3)2−3=9−3=6 (−3, 6)(−3, 6)
−2−2 y=(−2)2−3=4−3=1y=(−2)2−3=4−3=1 (−2, 1)(−2, 1)
−1−1 y=(−1)2−3=1−3=−2y=(−1)2−3=1−3=−2 (−1, −2)(−1, −2)
0 y=02−3=0−3=−3y=02−3=0−3=−3 (0, −3)(0, −3)
1 y=12−3=1−3=−2y=12−3=1−3=−2 (1, −2)(1, −2)
2 y=22−3=4−3=1y=22−3=4−3=1 (2, 1)
3 y=32−3=9−3=6y=32−3=9−3=6 (3, 6)

We plot the seven solutions (x, y) and join them with a smooth curve, as shown in Figure 2.11. This curve is the graph of the equation y=x2−3.y=x2−3.

Figure 2.11

Graph of a parabola

Practice Problem 1

  1. Sketch the graph of y=−x2+1.y=−x2+1.

The bowl-shaped curve sketched in Figure 2.11 is called a parabola. It is easy to find parabolas in everyday settings. For example, when you throw a ball, the path it travels is a parabola. Also, the reflector behind a car’s headlight is parabolic in shape.

Example 1 suggests the following three steps for sketching the graph of an equation by plotting points.

Comment

This point-plotting technique has obvious pitfalls. For instance, many different curves pass through the four points. See Figure 2.12. Assume that these points are solutions of a given equation. There is no way to guarantee that any curve we pass through the plotted points is the actual graph of the equation. However, in general, more plotted solutions result in a more accurate graph of the equation.

Figure 2.12

Several graphs through four points

We now discuss some special features of the graph of an equation.

Intercepts

  1. 2 Find the intercepts of a graph.

We examine the points where a graph intersects (crosses or touches) the coordinate axes. Because all points on the x-axis have a y-coordinate of 0, any point where a graph intersects the x-axis has the form (a, 0). See Figure 2.13. The number a is called an x-intercept of the graph. Similarly, any point where a graph intersects the y-axis has the form (0, b), and the number b is called a y-intercept of the graph.

Figure 2.13

Intercepts of a graph

Side Note

Do not try to calculate the x-intercept by setting x=0.x=0. The term x-intercept denotes the x-coordinate of the point where the graph touches or crosses the x-axis; so the y-coordinate must be 0.

Example 2 Finding Intercepts

Find the x- and y-intercepts of the graph of the equation y=x2−x−2.y=x2−x−2.

Solution

  1. Step 1 Set y=0y=0 in the equation and solve for x.

    y=x2−x−2Original equation0=x2−x−2Set y=0.0=(x+1)(x−2)Factor.x+1=0orx−2=0Zero-product propertyx=−1orx=2Solve each equation for x.
    y00x+1x=====x2−x−2x2−x−2(x+1)(x−2)0orx−2=0−1orx=2Original equationSet y=0.Factor.Zero-product propertySolve each equation for x.

    The x-intercepts are −1−1 and 2.

  2. Step 2 Set x=0x=0 in the equation and solve for y.

    y=x2−x−2Original equationy=02−0−2Set x=0.y=−2Solve for y.
    y=x2−x−2y=02−0−2y=−2Original equationSet x=0.Solve for y.

    The y-intercept  is −2.−2.

    The graph of the equation y=x2−x−2y=x2−x−2  is shown in Figure 2.14 .

    Figure 2.14

    Intercepts of a graph

Practice Problem 2

  1. Find the intercepts of the graph of y=2x2+3x−2.y=2x2+3x−2.

Symmetry

  1. 3 Find the symmetries in a graph.

The concept of symmetry helps us sketch graphs of equations. A graph has symmetry if one portion of the graph is a mirror image of another portion. As shown in Figure 2.15(a), if a line ℓℓ is an axis of symmetry, or line of symmetry, we can construct the mirror image of any point P not on ℓℓ by first drawing the perpendicular line segment from P to ℓℓ. Then we extend this segment an equal distance on the other side to a point P′P′ so that the line ℓℓ perpendicularly bisects the line segment ¯PP′PP′¯¯¯¯¯¯. In Figure 2.15(a), we say that the point P′P′ is the symmetric image of the point P about the line ℓℓ.

Figure 2.15

Two points M and M′M′ are symmetric about a point Q if Q is the midpoint of the line segment ¯MM′.MM′¯¯¯¯¯¯¯¯. Figure 2.15(b) illustrates the symmetry about the origin O. Symmetry lets us use information about part of the graph to draw the remainder of the graph.

The following three types of symmetries are frequently used.

Example 3 Checking for Symmetry

Determine whether the graph of the equation y=1x2+5y=1x2+5 is symmetric about the y-axis.

Solution

Replace x with −x−x to see if (−x, y)(−x, y) also satisfies the equation.

y=1x2+5Original equationy=1(−x)2+5Replace x with −x.y=1x2+5Simplify: (−x)2=x2.
y=1x2+5y=1(−x)2+5y=1x2+5Original equationReplace x with −x.Simplify: (−x)2=x2.

Because replacing x with −x−x gives us the original equation, the graph of y=1x2+5y=1x2+5 is symmetric with respect to the y-axis.

Practice Problem 3

  1. Check whether the graph of x2−y2=1x2−y2=1 is symmetric about the y-axis.

Side Note

Note that if only even powers of x appear in an equation, then the graph is symmetric with respect to the y-axis because for any integer n, (−x)2n=x2n(−x)2n=x2n.

Practice Problem 4

  1. Check whether the graph of x2=y3x2=y3 is symmetric about the x-axis, the y-axis, and the origin.

Example 5 Sketching a Graph by Using Intercepts and Symmetry

Initially, 400 deer are on Leafs island. The number y of deer on the island after t years is described by the equation

y=−t4+96t2+400.
y=−t4+96t2+400.

  1. Sketch the graph of the equation y=−t4+96t2+400.y=−t4+96t2+400.

  2. Adjust the graph in part a to account for only the physical aspects of the problem.

  3. When do deer become extinct on Leafs?

Solution

    1. We find all intercepts. If we set t=0t=0 in the equation y=−t4+96t2+400,y=−t4+96t2+400, we obtain y=400.y=400. Thus, the y-intercept is 400.

      To find the t-intercepts, we set y=0y=0 in the given equation.

      y=−t4+96t2+400Original equation0=−t4+96t2+400Set y=0.t4−96t2−400=0Multiply both sides by −1and interchange sides.(t2+4)(t2−100)=0Factor.(t2+4)(t+10)(t−10)=0Factor t2−100.t2+4=0ort+10=0ort−10=0Zero-product propertyt=−10ort=10Solve for t; there is no realsolution of t2+4=0.
      y0t4−96t2−400(t2+4)(t2−100)(t2+4)(t+10)(t−10)t2+4=0ort+10t=======−t4+96t2+400−t4+96t2+4000000ort−10=0−10ort=10Original equationSet y=0.Multiply both sides by −1and interchange sides.Factor.Factor t2−100.Zero-product propertySolve for t; there is no realsolution of t2+4=0.

      The t-intercepts are −10−10 and 10.

    2. We check for symmetry. Note that t replaces x as the independent variable.

      Symmetry about the t-axis:

      Replacing y with −y−y gives −y=−t4+96t2+400.−y=−t4+96t2+400. The pair (0, 400) is on the graph of y=−t4+96t2+400y=−t4+96t2+400 but not on −y=−t4+96t2+400−y=−t4+96t2+400. Consequently, the graph is not symmetric about the t-axis.

      Symmetry about the y-axis:

      Replacing t with −t−t in the equation y=−t4+96t2+400,y=−t4+96t2+400, we obtain y=−(−t)4+96(−t)2+400=−t4+96t2+400,y=−(−t)4+96(−t)2+400=−t4+96t2+400, which is the original equation. Thus, (−t, y)(−t, y) also satisfies the equation and the graph is symmetric about the y-axis.

      Symmetry about the origin:

      Replacing t with −t−t and y with −y−y in the equation y=−t4+96t2+400,y=−t4+96t2+400, we obtain −y=−(−t)4+96(−t)2+400, or −y=−t4+96t2+400.−y=−(−t)4+96(−t)2+400, or −y=−t4+96t2+400. As we saw when we discussed symmetry in the t-axis, (0, 400) is a solution but (0,−400)(0,−400) is not a solution of this equation; so the graph is not symmetric with respect to the origin.

    3. We sketch the graph by plotting points for t≥0t≥0 (see Table 2.4) and then using symmetry about the y-axis (see Figure 2.17).

      Figure 2.17

      Table 2.4

      t y=−t4+96 t2+400y=−t4+96 t2+400 (t, y)
      0 400 (0, 400)
      1 495 (1, 495)
      5 2175 (5, 2175)
      7 2703 (7, 2703)
      9 1615 (9, 1615)
      10 0 (10, 0)
      11 −2625−2625 (11, −2625)(11, −2625)

  2. The graph pertaining to the physical aspects of the problem is the red portion of the graph in Figure 2.17.

  3. The positive t-intercept, 10, gives the time in years when the deer population of Leafs is 0; so deer are extinct after 10 years.

Practice Problem 5

  1. Repeat Example 5 , assuming that the initial deer population is 324 and the number of deer on the island after t years is given by the equation y=−t4+77t2+324.y=−t4+77t2+324.

Circles

  1. 4 Find the equation of a circle.

Sometimes a curve that is described geometrically can also be described by an algebraic equation. We illustrate this situation in the case of a circle.

Standard Form

A point P(x, y) is on the circle if and only if its distance from the center C(h, k) is r. Using the notation for the distance between the points P and C, we have

d(P, C)=r√(x−h)2+(y−k)2=rDistance formula(x−h)2+(y−k)2=r2Square both sides.
d(P, C)(x−h)2+(y−k)2−−−−−−−−−−−−−−−√(x−h)2+(y−k)2===rrr2Distance formulaSquare both sides.

The equation (x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2 is an equation of a circle with radius r and center (h, k). A point (x, y) is on the circle of radius r with center C(h, k) if and only if it satisfies this equation. Figure 2.18 is the graph of a circle with center C(h, k) and radius r.

Figure 2.18

Example 6 Finding the Equation of a Circle

Find the standard form of the equation of the circle with center (7,−3)(7,−3) and that passes through the point P=(5,−2).P=(5,−2).

Solution

(x−h)2+(y−k)2=r2Standard form(x−7)2+(y−(−3))2=r2Replace h with 7 and k with −3.(x−7)2+(y+3)2=r2−(−3)=3
(x−h)2+(y−k)2(x−7)2+(y−(−3))2(x−7)2+(y+3)2===r2r2r2Standard formReplace h with 7 and k with −3.−(−3)=3
(2)

Because the point P=(5,−2)P=(5,−2) lies on the circle, its coordinates satisfy equation (2). So

(5−7)2+(−2+3)2=r2Replace x with 5 and y with −2.5=r2Simplify.
(5−7)2+(−2+3)2=r25=r2Replace x with 5 and y with −2.Simplify.

Replacing r2r2 with 5 in equation (2) gives the required standard form

(x−7)2+(y+3)2=5.
(x−7)2+(y+3)2=5.

Practice Problem 6

  1. Find the standard form of the equation of the circle with center (3,−6)(3,−6) and radius 10.

If an equation in two variables can be written in standard form (1), then its graph is a circle with center (h, k) and radius r.

Example 7 Graphing a Circle

Specify the center and radius and graph each circle.

  1. x2+y2=1x2+y2=1

  2. (x+2)2+(y−3)2=25(x+2)2+(y−3)2=25

Solution

  1. The equation x2+y2=1x2+y2=1 can be rewritten as

    (x−0)2+(y−0)2=12.
    (x−0)2+(y−0)2=12.

    Comparing this equation with equation (1), we conclude that the given equation is an equation of the circle with center (0, 0) and radius 1. The graph is shown in Figure 2.19. This circle is called the unit circle.

    Figure 2.19

    The unit circle

  2. Rewriting the equation (x+2)2+(y−3)2=25(x+2)2+(y−3)2=25 as

    [x−(−2)]2+(y−3)2=52,x+2=x−(−2)
    [x−(−2)]2+(y−3)2=52,x+2=x−(−2)

    we see that the graph of this equation is the circle with center (−2, 3)(−2, 3) and radius 5. The graph is shown in Figure 2.20.

    Figure 2.20

    Circle with radius 5 and center at (−2, 3)(−2, 3)

Practice Problem 7

  1. Graph the equation (x−2)2+(y+1)2=36.(x−2)2+(y+1)2=36.

Semicircles

Letting h=0h=0 and k=0k=0 in equation (1), we have

x2+y2=r2
x2+y2=r2
(3)

Equation (3) is the standard form for a circle with center at the origin and radius r.

Side Note

If r≥x≥0r≥x≥0, then r2≥x2r2≥x2, so r2−x2≥0r2−x2≥0.

If −r≤−x≤0−r≤−x≤0, then r≥x≥0r≥x≥0, and again r2−x2≥0r2−x2≥0. So if −r≤x≤r−r≤x≤r, we have r2−x2≥0r2−x2≥0.

We solve equation (3) for y:

y2=r2−x2Subtract x2 from both sides.y=±√r2−x2Square root property.
y2y==r2−x2±r2−x2−−−−−−√Subtract x2 from both sides.Square root property.

or

y=√r2+x2andy=√r2−x2.
y=r2+x2−−−−−−√andy=r2−x2−−−−−−√.

Similarly, solving equation (3) for x, we have two equations,

x=√r2−y2 and  x=−√r2−y2.
x=r2−y2−−−−−−√ and  x=−r2−y2−−−−−−√.

The graphs of these four equations are semicircles (half circles), shown in Figure 2.21.

Figure 2.21

Semicircles

General Form

If we expand the squared expressions in the standard equation of a circle,

(x−h)2+(y−k)2=r2,
(x−h)2+(y−k)2=r2,

and then simplify, we obtain an equation of the form

x2+y2+ax+by+c=0.
x2+y2+ax+by+c=0.
(4)

Equation (4) is called the general form of the equation of a circle. An equation Ax2+By2+Cx+Dy+E=0Ax2+By2+Cx+Dy+E=0 with A≠0A≠0 and A=BA=B can be converted to the general form of the equation of a circle by dividing both sides by A. See Exercise 91.

On the other hand, if we are given an equation in general form, we can convert it to standard form by completing the squares on the x- and y-terms. This gives

(x−h)2+(y−k)2=d.
(x−h)2+(y−k)2=d.
(5)

If d>0,d>0, the graph of equation (5) is a circle with center (h, k) and radius √d.d−−√. If d=0,d=0, the graph of equation (5) is the point (h, k). If d<0,d<0, there is no graph.

Example 8 Converting the General Form to Standard Form

Find the center and radius of the circle with equation

x2+y2−6x+8y+10=0.
x2+y2−6x+8y+10=0.

Solution

Complete the squares on both the x-terms and y-terms to get standard form.

x2+y2−6x+8y+10=0Original equation(x2−6x)+(y2+8y)=−10Group the x-termsand y-terms.(x2−6x+9)+(y2+8y+16)=−10+9+16Complete the squaresby adding 9 and 16 toboth sides.(x−3)2+(y+4)2=15Factor and simplify.(x−3)2+[y−(−4)]2=(√15)2
x2+y2−6x+8y+10(x2−6x)+(y2+8y)(x2−6x+9)+(y2+8y+16)(x−3)2+(y+4)2(x−3)2+[y−(−4)]2=====0−10−10+9+1615(15−−√)2Original equationGroup the x-termsand y-terms.Complete the squaresby adding 9 and 16 toboth sides.Factor and simplify.

The last equation tells us that we have h=3, k=−4,h=3, k=−4, and r=√15.r=15−−√. Therefore, the circle has center (3,−4)(3,−4) and radius √15≈3.9.15−−√≈3.9.

Practice Problem 8

  1. Find the center and radius of the circle with equation x2+y2+4x−6y−12=0.x2+y2+4x−6y−12=0.

Section 2.2 Exercises

Concepts and Vocabulary

  1. The graph of an equation in two variables such as x and y is the set of all ordered pairs (a, b)                   .

  2. If (−2, 4)(−2, 4) is a point on a graph that is symmetric with respect to the y-axis, then the point                    is also on the graph.

  3. If (0,−5)(0,−5) is a point of a graph, then −5−5 is a(n)                     intercept of the graph.

  4. An equation in standard form of a circle with center (1, 0) and radius 2 is                   .

  5. True or False. The graph of the equation 3x2−2x+y+3=03x2−2x+y+3=0 is a circle.

  6. True or False. If a graph is symmetric about the x-axis, then it must have at least one x-intercept.

  7. True or False. The center of the circle with equation (x+3)2+(y+4)2=9(x+3)2+(y+4)2=9 is the point (3, 4).

  8. True or False. If (−2, 3)(−2, 3) is a point on a graph that is symmetric with respect to the origin, then the point (2,−3)(2,−3) is also on the graph.

Building Skills

In Exercises 9–14, determine whether the given points are on the graph of the equation.

Equation Points
9. y=x−1y=x−1 (−3,−4), (1, 0), (4, 3), (2, 3)(−3,−4), (1, 0), (4, 3), (2, 3)
10. 2y=3x+52y=3x+5 (−1, 1), (0, 2),(−53, 0), (1, 4)(−1, 1), (0, 2),(−53, 0), (1, 4)
11. y=√x+1y=x+1−−−−−√ (3, 2), (0, 1), (8,−3), (8, 3)(3, 2), (0, 1), (8,−3), (8, 3)
12. y=1xy=1x (−3, 13), (1, 1), (0, 0),(2, 12)(−3, 13), (1, 1), (0, 0),(2, 12)
13. x2−y2=1x2−y2=1 (1, 0), (0,−1), (2, √3), (2,−√3)(1, 0), (0,−1), (2, 3–√), (2,−3–√)
14. y2=xy2=x (1,−1), (1, 1) (0, 0), (2,−√2)(1,−1), (1, 1) (0, 0), (2,−2–√)

In Exercises 15–36, graph each equation by plotting points. Let x=−3, −2, −1, 0, 1, 2,x=−3, −2, −1, 0, 1, 2, and 3 where applicable.

  1. y=x+1y=x+1

  2. y=x−1y=x−1

  3. y=2xy=2x

  4. y=12xy=12x

  5. y=x2y=x2

  6. y=−x2y=−x2

  7. y=|x|y=|x|

  8. y=|x+1|y=|x+1|

  9. y=|x|+1y=|x|+1

  10. y=−|x|+1y=−|x|+1

  11. y=4−x2y=4−x2

  12. y=x2−4y=x2−4

  13. y=√9−x2y=9−x2−−−−−√

  14. y=−√9−x2y=−9−x2−−−−−√

  15. y=x3y=x3

  16. y=−x3y=−x3

  17. y3=xy3=x

  18. y3=−xy3=−x

  19. x=|y|x=|y|

  20. |x|=|y||x|=|y|

  21. y=|2−x|y=|2−x|

  22. |x|+|y|=1|x|+|y|=1

In Exercises 37–46, find

  1. x and y-intercepts.

  2. symmetries (if any) about the x-axis, the y-axis, and the origin.

In Exercises 47–50, complete the given graph so that it has the indicated symmetry.

  1. symmetry about the x-axis

  2. symmetry about the y-axis

  3. symmetry about the origin

  4. symmetry about the x-axis and symmetry about the y-axis

In Exercises 51–64, find the x- and y-intercepts of the graph of each equation (if any).

  1. 3x+4y=123x+4y=12

  2. 2x+3y=52x+3y=5

  3. x5+y3=1x5+y3=1

  4. x2−y3=1x2−y3=1

  5. y=x+2x−1y=x+2x−1

  6. x=y−2y+1x=y−2y+1

  7. y=x2−6x+8y=x2−6x+8

  8. x=y2−5y+6x=y2−5y+6

  9. x2+y2=4x2+y2=4

  10. (x−1)2+y2=9(x−1)2+y2=9

  11. y=√9−x2y=9−x2−−−−−√

  12. y=√x2−1y=x2−1−−−−−√

  13. xy=1xy=1

  14. y=x2+1y=x2+1

In Exercises 65–74, test each equation for symmetry with respect to the x-axis, the y-axis, and the origin.

  1. y=x2+1y=x2+1

  2. x=y2+1x=y2+1

  3. y=x3+xy=x3+x

  4. y=2x3−xy=2x3−x

  5. y=5x4+2x2y=5x4+2x2

  6. y=−3x6+2x4+x2y=−3x6+2x4+x2

  7. y=−3x5+2x3y=−3x5+2x3

  8. y=2x2−|x|y=2x2−∣∣x∣∣

  9. x2y2+2xy=1x2y2+2xy=1

  10. x2+y2=16x2+y2=16

In Exercises 75–78, specify the center and the radius of each circle.

  1. (x−2)2+(y−3)2=36(x−2)2+(y−3)2=36

  2. (x+1)2+(y−3)2=16(x+1)2+(y−3)2=16

  3. (x+2)2+(y+3)2=11(x+2)2+(y+3)2=11

  4. (x−12)2+(y+32)2=34(x−12)2+(y+32)2=34

In Exercises 79–88, find the standard form of the equation of a circle that satisfies the given conditions. Graph each equation.

  1. Center (0, 1);(0, 1); radius 2

  2. Center (1, 0);(1, 0); radius 1

  3. Center (−1, 2);(−1, 2); radius √22–√

  4. Center (−2,−3);(−2,−3); radius √77–√

  5. Center (3,−4);(3,−4); passing through the point (−1, 5)(−1, 5)

  6. Center (−1, 1);(−1, 1); passing through the point (2, 5)

  7. Center (1, 2); touching the x-axis

  8. Center (1, 2); touching the y-axis

  9. Diameter with endpoints (7, 4) and (−3, 6)(−3, 6)

  10. Diameter with endpoints (2,−3)(2,−3) and (8, 5)

In Exercises 89–94, find

  1. the center and radius of each circle.

  2. the x- and y-intercepts of the graph of each circle.

  1. x2+y2−2x−2y−4=0x2+y2−2x−2y−4=0

  2. x2+y2−4x−2y−15=0x2+y2−4x−2y−15=0

  3. 2x2+2y2+4y=02x2+2y2+4y=0

  4. 3x2+3y2+6x=03x2+3y2+6x=0

  5. x2+y2−x=0x2+y2−x=0

  6. x2+y2+1=0x2+y2+1=0

Applying the Concepts

In Exercises 95 and 96, a graph is described geometrically as the path of a point P(x, y) on the graph. Find an equation for the graph described.

  1. Geometry. P(x, y) is on the graph if and only if the distance from P(x, y) to the x-axis is equal to its distance to the y-axis.

  2. Geometry. P(x, y) is the same distance from the two points (1, 2) and (3,−4)(3,−4).

  3. Saving and spending. Sketch a graph (years/ money) that shows the amount of money available to you if you save $100 each month until you have $2400 and then withdraw $80 each month until the $2400 is gone.

  4. Tracking distance during a workout. Sketch a graph (minutes/miles) that shows the distance you have traveled from your starting point if you jog at 6 mph for 10 minutes, rest for 10 minutes, and then walk at 3 mph back to your starting point.

  5. Corporate profits. The equation P=−0.5t2−3t+8P=−0.5t2−3t+8 describes the monthly profits (in millions of dollars) of ABCD Corp. for the year 2018, with t=0t=0 representing July 2018.

    1. How much profit did the corporation make in March 2018?

    2. How much profit did the corporation make in October 2018?

    3. Sketch the graph of the equation.

    4. Find the t-intercepts. What do they represent?

    5. Find the P-intercept. What does it represent?

  6. Female students in colleges. The equation

    P=−0.002t2+0.51t+17.5
    P=−0.002t2+0.51t+17.5

    models the approximate number (in millions) of female college students in the United States for the academic years 2005–2009, with t=0t=0 representing 2005.

    1. Sketch the graph of the equation.

    2. Find the P-intercept. What does it represent?

      (Source: Statistical Abstracts of the United States)

  7. Motion. An object is thrown up from the top of a building that is 320 feet high. The equation y=−16t2+128t+320y=−16t2+128t+320 gives the object’s height (in feet) above the ground at any time t (in seconds) after the object is thrown.

    1. What is the height of the object after 0, 1, 2, 3, 4, 5, and 6 seconds?

    2. Sketch the graph of the equation y=−16t2+128t+320.y=−16t2+128t+320.

    3. What part of the graph represents the physical aspects of the problem?

    4. What are the intercepts of this graph, and what do they mean?

  8. Diving for treasure. A treasure-hunting team of divers is placed in a computer-controlled diving cage. The equation d=403t−29t2d=403t−29t2 describes the depth d (in feet) that the cage will descend in t minutes.

    1. Sketch the graph of the equation d=403t−29t2.d=403t−29t2.

    2. What part of the graph represents the physical aspects of the problem?

    3. What is the total time of the entire diving experiment?

Beyond the Basics

  1. In the same coordinate system, sketch the graphs of the two circles with equations x2+y2−4x+2y−20=0x2+y2−4x+2y−20=0 and x2+y2−4x+2y−31=0 and find the area of the region bounded by the two circles.

  2. Find the equation of a circle with radius 5 and x-intercepts −4 and 4.

    [Hint: Center must be on the y-axis; there are two such circles.]

Critical Thinking / Discussion / Writing

  1. Sketch the graph of y2=2x and explain how this graph is related to the graphs of y=√2x and y=−√2x.

  2. Show that a graph that is symmetric with respect to the x-axis and the y-axis must also be symmetric with respect to the origin. Give an example to show that the converse is not true.

    1. Show that a circle with diameter having endpoints A(0, 1) and B(6, 8) intersects the x-axis at the roots of the equation x2−6x+8=0.

    2. Show that a circle with diameter having endpoints A(0, 1) and B(a, b) intersects the x-axis at the roots of the equation x2−ax+b=0.

    3. Use graph paper, ruler, and compass to approximate the roots of the equation x2−3x+1=0.

  4. The figure shows two circles each with radius r.

    1. Write the coordinates of the center of each circle.

    2. Find the area of the shaded region.

Getting Ready for the Next Section

In Exercises 109–114, perform the indicated operations.

  1. 5−36−2

  2. 1−2−2−2

  3. 2−(−3)3−13

  4. 3−1−2−(−6)

  5. 12−1438−(−14)

  6. 34−112−16

In Exercises 115–120, write the negative reciprocal of each number.

  1. 2

  2. −3

  3. −23

  4. 43

  5. 1−122+34

  6. 23−1456−(−34)

In Exercises 121–124, solve each equation for the specified variable.

  1. 2x+3y=6 for y

  2. x2−y5=3 for y

  3. y−2−23 (x+1)=0 for y

  4. 0.1x+0.2y=0 for y

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