7.1 Graphing and Functions - Moreland MATH · 7 Algebra: Graphs, Functions and Linear Systems > 7.1 Graphing and Functions 7.1 Graphing and Functions What am I Supposed to Learn?

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7 Algebra: Graphs, Functions and Linear Systems > 7.1 Graphing and Functions

7.1 Graphing and Functions

What am I Supposed to Learn?After you have read this section, you should be able to:

1 Plot points in the rectangular coordinate system.

2 Graph equations in the rectangular coordinate system.

3 Use function notation.

4 Graph functions.

5 Use the vertical line test.

6 Obtain information about a function from its graph.

THE BEGINNING OF THE SEVENTEENTH CENTURY was a time of innovative ideas and enormous intellectual progress in Europe. English theatergoers enjoyed asuccession of exciting new plays by Shakespeare. William Harvey proposed the radical notion that the heart was a pump for blood rather than the center of emotion.Galileo, with his new-fangled invention called the telescope, supported the theory of Polish astronomer Copernicus that the sun, not the Earth, was the center of thesolar system. Monteverdi was writing the world's first grand operas. French mathematicians Pascal and Fermat invented a new field of mathematics called probabilitytheory.

Into this arena of intellectual electricity stepped French aristocrat René Descartes (1596–1650). Descartes (pronounced “day cart”), propelled by the creativitysurrounding him, developed a new branch of mathematics that brought together algebra and geometry in a unified way—a way that visualized numbers as points on agraph, equations as geometric figures, and geometric figures as equations. This new branch of mathematics, called analytic geometry, established Descartes as one ofthe founders of modern thought and among the most original mathematicians and philosophers of any age. We begin this section by looking at Descartes's deceptivelysimple idea, called the rectangular coordinate system or (in his honor) the Cartesian coordinate system.

Points and Ordered Pairs

1 Plot points in the rectangular coordinate system.

Descartes used two number lines that intersect at right angles at their zero points, as shown in Figure 7.1. The horizontal number line is the x-axis. The vertical numberline is the y-axis. The point of intersection of these axes is their zero points, called the origin. Positive numbers are shown to the right and above the origin. Negativenumbers are shown to the left and below the origin. The axes divide the plane into four quarters, called quadrants. The points located on the axes are not in anyquadrant.

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Page 408

dFIGURE 7.1 The rectangular coordinate system

Each point in the rectangular coordinate system corresponds to an ordered pair of real numbers, Examples of such pairs are and The firstnumber in each pair, called the x-coordinate, denotes the distance and direction from the origin along the x-axis. The second number in each pair, called the y-coordinate, denotes vertical distance and direction along a line parallel to the y-axis or along the y-axis itself.

Figure 7.2 shows how we plot, or locate, the points corresponding to the ordered pairs and We plot by going 5 units from 0 to the left alongthe x-axis. Then we go 3 units up parallel to the y-axis. We plot by going 3 units from 0 to the right along the x-axis and 5 units down parallel to the y-axis. Thephrase “the points corresponding to the ordered pairs and ” is often abbreviated as “the points and ”

dFIGURE 7.2 Plotting and

Great Question!What's the significance of the word ordered when describing a pair of real numbers

The phrase ordered pair is used because order is important. The order in which coordinates appear makes a difference in a point's location. This is illustrated inFigure 7.2.

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(x, y). (−5, 3) (3, −5).

(−5, 3) (3, −5). (−5, 3)(3, −5)

(−5, 3) (3, −5) (−5, 3) (3, −5).

(−5, 3) (3, −5)



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7 Algebra: Graphs, Functions and Linear Systems > 7.1 Graphing and Functions > Graphs of Equations

Example 1 Plotting Points in the Rectangular Coordinate SystemPlot the points: C(5, 0), E(0, 4), and F(0, 0).

SOLUTION

See Figure 7.3. We move from the origin and plot the points in the following way:

3 units left, 5 units up2 units right, 4 units down

C(5, 0): 5 units right, 0 units up or down5 units left, 3 units down

E(0, 4): 0 units right or left, 4 units up

0 units right or left, 0 units up or down

FIGURE 7.3 Plotting points

Check Point 1Plot the points: and

Graphs of Equations

2 Graph equations in the rectangular coordinate system.

A relationship between two quantities can sometimes be expressed as an equation in two variables, such as

A solution of an equation in two variables, x and y, is an ordered pair of real numbers with the following property: When the x-coordinate is substituted for x and they-coordinate is substituted for y in the equation, we obtain a true statement. For example, consider the equation and the ordered pair When 3 issubstituted for x and is substituted for y, we obtain the statement Because this statement is true, the ordered pair

is a solution of the equation We also say that satisfies the equation.

We can generate as many ordered-pair solutions as desired to by substituting numbers for x and then finding the corresponding values for y. For example,suppose we let



A(−3, 5), B(2, −4), D(−5, −3),

A(−3, 5):

B(2, −4):

D(−5, −3):

F (0, 0):

▲

Notice that the origin

is represented by(0, 0).

A(−2, 4), B(4, −2), C(−3, 0), D(0, −3).

y = 4 − .x2

y = 4 − x2 (3, −5).−5 −5 = 4 − , or − 5 = 4 − 9, or − 5 = −5.32

(3, −5) y = 4 − .x2 (3, −5)

y = 4 − x2

x = 3:

Start with x.

▼

x

Compute y.

▼

y = 4 − x2

Form the ordered pair (x, y).

▼

Ordered Pair(x, y)

y = 4 − = 4 − 9 = −532








Page 409

The graph of an equation in two variables is the set of all points whose coordinates satisfy the equation. One method for graphing such equations is the point-plotting method. First, we find several ordered pairs that are solutions of the equation. Next, we plot these ordered pairs as points in the rectangular coordinate system.Finally, we connect the points with a smooth curve or line. This often gives us a picture of all ordered pairs that satisfy the equation.


3

▲

Let x = 3.

(3, −5)

▲

(3, −5) is a solution of y = 4 − .x2








7 Algebra: Graphs, Functions and Linear Systems > 7.1 Graphing and Functions > Graphs of Equations

Example 2 Graphing an Equation Using the Point-Plotting MethodGraph Select integers for x, starting with and ending with 3.

SOLUTION

For each value of x, we find the corresponding value for y.

0 (0, 4)

1 (1, 3)

2 (2, 0)

3

Now we plot the seven points and join them with a smooth curve, as shown in Figure 7.4. The graph of is a curve where the part of the graph to the rightof the y-axis is a reflection of the part to the left of it and vice versa. The arrows on the left and the right of the curve indicate that it extends indefinitely in bothdirections.

FIGURE 7.4 The graph of

Check Point 2Graph Select integers for x, starting with and ending with 3.

Part of the beauty of the rectangular coordinate system is that it allows us to “see” formulas and visualize the solution to a problem. This idea is demonstrated inExample 3.

Example 3 An Application Using Graphs of EquationsThe toll to a bridge costs $2.50. Commuters who use the bridge frequently have the option of purchasing a monthly discount pass for $21.00. With the discount pass,the toll is reduced to $1.00. The monthly cost, y, of using the bridge x times can be described by the following formulas:

Without the discount pass:

With the discount pass:

a. Let 2, 4, 10, 12, 14, and 16. Make a table of values for each equation showing seven solutions for the equation.



y = 4 − .x2 −3

Start with x.

▼

x

Compute y.

▼

y = 4 − x2

Form the ordered pair (x, y).

▼

Ordered Pair(x, y)

►

We selected integers

from − 3 to 3, inclusive,

to include three negative

numbers, 0, and three

positive numbers. We

also wanted to keep the

resulting computations

for y relatively simple.

−3 y = 4 − = 4 − 9 = −5(−3)2 (−3, −5)

−2 y = 4 − = 4 − 4 = 0(−2)2 (−2, 0)

−1 y = 4 − = 4 − 1 = 3(−1)2 (−1, 3)

y = 4 − = 4 − 0 = 402

y = 4 − = 4 − 1 = 312

y = 4 − = 4 − 4 = 022

y = 4 − = 4 − 9 = −532 (3, −5)

y = 4 − x2

y = 4 − x2

y = 4 − x. −3

y = 2.5x

The monthly cost, y, is $2.50 times the number of times, x, that

the bridge is used.

y = 21 + 1 ⋅ x

y = 21 + x.

The monthly cost, y, is $21 for the discount pass plus $1 times

the number of times, x, that the bridge is used.

x = 0,








Page 410

b. Graph the equations in the same rectangular coordinate system.

c. What are the coordinates of the intersection point for the two graphs? Interpret the coordinates in practical terms.









7 Algebra: Graphs, Functions and Linear Systems > 7.1 Graphing and Functions > Functions

SOLUTION

a. Tables of values showing seven solutions for each equation follow.

WITHOUT THE DISCOUNTPASS

x (x, y)0 (0, 0)2 (2, 5)4 (4, 10)10 (10, 25)12 (12, 30)14 (14, 35)16 (16, 40)WITH THE DISCOUNT PASS

x (x, y)0 (0, 21)2 (2, 23)4 (4, 25)10 (10, 31)12 (12, 33)14 (14, 35)16 (16, 37)

b. Now we are ready to graph the two equations. Because the x- and y-coordinates are nonnegative, it is only necessary to use the origin, the positive portions ofthe x- and y-axes, and the first quadrant of the rectangular coordinate system. The x-coordinates begin at 0 and end at 16. We will let each tick mark on the x-axisrepresent two units. However, the y-coordinates begin at 0 and get as large as 40 in the formula that describes the monthly cost without the coupon book. So thatour y-axis does not get too long, we will let each tick mark on the y-axis represent five units. Using this setup and the two tables of values, we construct the graphsof and shown in Figure 7.5.

dFIGURE 7.5 Options for a toll

c. The graphs intersect at (14, 35). This means that if the bridge is used 14 times in a month, the total monthly cost without the discount pass is the same as thetotal monthly cost with the discount pass, namely $35.

In Figure 7.5, look at the two graphs to the right of the intersection point (14, 35). The red graph of lies below the blue graph of This meansthat if the bridge is used more than 14 times in a month the (red) monthly cost, y, with the discount pass is less than the (blue) monthly cost, y, without thediscount pass.

Check Point 3The toll to a bridge costs $2.00. If you use the bridge x times in a month, the monthly cost, y, is With a $10 discount pass, the toll is reduced to $1.00. Themonthly cost, y, of using the bridge x times in a month with the discount pass is

a. Let Make tables of values showing seven solutions of and seven solutions of

b. Graph the equations in the same rectangular coordinate system.

c. What are the coordinates of the intersection point for the two graphs? Interpret the coordinates in practical terms.



y = 2.5x

y = 2.5(0) = 0

y = 2.5(2) = 5

y = 2.5(4) = 10

y = 2.5(10) = 25

y = 2.5(12) = 30

y = 2.5(14) = 35

y = 2.5(16) = 40

y = 21+ x

y = 21 + 0 = 21

y = 21 + 2 = 23

y = 21 + 4 = 25

y = 21 + 10 = 31

y = 21 + 12 = 33

y = 21 + 14 = 35

y = 21 + 16 = 37

y = 2.5x y = 21 + x,

y = 21 + x y = 2.5x.(x > 14),

y = 2x.y = 10 + x.

x = 0, 2, 4, 6, 8, 10, and 12. y = 2x y = 10 + x.








Page 411

Functions

3 Use function notation.

Reconsider one of the equations from Example 3, Recall that this equation describes the monthly cost, y, of using the bridge x times, with a toll cost of $2.50each time the bridge is used. The monthly cost, y, depends on the number of times the bridge is used, x. For each value of x, there is one and only one value of y. If anequation in two variables (x and y) yields precisely one value of y for each value of x, we say that y is a function of x.

The notation indicates that the variable y is a function of x. The notation is read “f of x.”


y = 2.5x.

y = f(x) f(x)









For example, the formula for the cost of the bridge

can be expressed in function notation as

We read this as “f of x is equal to 2.5x.” If, say, x equals 10 (meaning that the bridge is used 10 times), we can find the corresponding value of y (monthly cost) using theequation

Because (f of 10 equals 25), this means that if the bridge is used 10 times in a month, the total monthly cost is $25.

Great Question!Doesn't f(x) indicate that I need to multiply f and x?

The notation does not mean “f times x.” The notation describes the “output” for the function f when the “input” is x. Think of as another name for y.

Table 7.1 compares our previous notation with the new notation of functions.

TABLE 7.1 Function Notation“y Equals” Notation “ Equals” Notation

In our next example, we will apply function notation to three different functions. It would be awkward to call all three functions f. We will call the first function f, the secondfunction g, and the third function h. These are the letters most frequently used to name functions.

Example 4 Using Function NotationFind each of the following:

a.

b.

c.

SOLUTION



y = 2.5x

f(x) = 2.5x.

f(x) = 2.5x.

f(x)

f(10)

= 2.5x

= 2.5(10)

= 25

To find f(10), read “f of 10, ” replace x with 10.

f(10) = 25

f(x) f(x)

f(x)

y = 2.5x f(x) = 2.5x

If x

y

= 10

= 2.5(10) = 25.

f (10) = 2.5 (10) = 25

f of 10 equals 25.

f(4) for f(x) = 2x + 3

g(−2) for g(x) = 2 − 1x2

h(−5) for h(r) = − 2 + 5.r3 r2

a. f (x)

f (4)

f (4)

= 2x + 3

= 2 ⋅ 4 + 3

= 8 + 3

= 11

▲

f of 4 is 11.

This is the given function.

To find f of 4, replace x with 4.

Multiply: 2 ⋅ 4 = 8.

Add.








Page 412


b. g(x)

g(−2)

g(−2)

= 2 − 1x2

= 2 − 1(−2)2

= 2(4) − 1

= 8 − 1

= 7

▲

g of − 2 is 7.

This is the given function.

To find g of − 2, replace x with − 2.

Evaluate the exponential expression: = 4.(−2)2

Multiply: 2(4) = 8.

Subtract.

c. h (r)

h (−5)

h (−5)

= − 2 + 5r3 r2

= − 2 + 5(−5)3 (−5)2

= −125 − 2 (25) + 5

= −125 − 50 + 5

= −170

▲

h of − 5 is − 170.

The function's name is h and r represents the

function's input.

To find h of − 5, replace each occurrence of r

with − 5.

Evaluate exponential expressions: = −125(−5)3

and = 25.(−5)2

Multiply: 2 (25) = 50.

−125 − 50 = −175 and − 175 + 5 = −170.









Check Point 4Find each of the following:

a.

b.

c.

Example 5 An Application Involving Function NotationTailgaters beware: If your car is going 35 miles per hour on dry pavement, your required stopping distance is 160 feet, or the width of a football field. At 65 miles perhour, the distance required is 410 feet, or approximately the length of one and one-tenth football fields. Figure 7.6 shows stopping distances for cars at variousspeeds on dry roads and on wet roads. Figure 7.7 uses a line graph to represent stopping distances at various speeds on dry roads.

dFIGURE 7.6

dFIGURE 7.7 Source: National Highway Traffic Safety Administration



f(6) for f(x) = 4x + 5

g(−5) for g(x) = 3 − 10x2

h(−4) for h(r) = − 7r + 2.r2








Page 413

a. Use the line graph in Figure 7.7 to estimate a car's required stopping distance at 60 miles per hour on dry pavement. Round to the nearest 10 feet.

b. The function

models a car's required stopping distance, in feet, on dry pavement traveling at x miles per hour. Use this function to find the required stopping distance at60 miles per hour. Round to the nearest foot.

SOLUTION

a. The required stopping distance at 60 miles per hour is estimated using the point shown in Figure 7.8. The second coordinate of this point extends slightly morethan midway between 300 and 400 on the vertical axis. Thus, 360 is a reasonable estimate. We conclude that at 60 miles per hour on dry pavement, the requiredstopping distance is approximately 360 feet.

FIGURE 7.8


f(x) = 0.0875 − 0.4x + 66.6x2

f(x),








7 Algebra: Graphs, Functions and Linear Systems > 7.1 Graphing and Functions > Graphing Functions

b. Now we use the given function to determine the required stopping distance at 60 miles per hour. We need to find f(60). The arithmetic gets somewhat “messy,” soit is probably a good idea to use a calculator.

We see that —that is, f of 60 is approximately 358. The model indicates that the required stopping distance on dry pavement at 60 miles per hour isapproximately 358 feet.

TechnologyOn most calculators, here is how to find

Many Scientific Calculators

Many Graphing Calculators

Check Point 5a. Use the line graph in Figure 7.7 on the previous page to estimate a car's required stopping distance at 40 miles per hour on dry pavement. Round to the nearestten feet.

b. Use the function in Example 5(b), to find the required stopping distance at 40 miles per hour. Round to the nearest foot.

Graphing Functions

4 Graph functions.

The graph of a function is the graph of its ordered pairs. In our next example, we will graph two functions.

Example 6 Graphing FunctionsGraph the functions and in the same rectangular coordinate system. Select integers for x from inclusive.

SOLUTION

For each function, we use the suggested values for x to create a table of some of the coordinates. These tables are shown below. Then, we plot the five points in eachtable and connect them, as shown in Figure 7.9. The graph of each function is a straight line. Do you see a relationship between the two graphs? The graph of g isthe graph of f shifted vertically up 4 units.



f(x)

f(60)

= 0.0875 − 0.4x + 66.6x2

= 0.0875 − 0.4(60) + 66.6(60)2

= 0.0875(3600) − 0.4(60) + 66.6

= 315 − 24 + 66.6

= 357.6

≈ 358

This function models stopping

distance, f(x), at x miles per hour.

Replace each x with 60.

Use the order of operations, first

evaluating the exponential expression.

Perform the multiplications.

Subtract and add as indicated.

As directed, we've rounded to the nearest foot.

f(60) ≈ 358

0.0875 − 0.4(60) + 66.6(60)2

.0875 60 .4 60 66.6× x2 − × + =

.0875 60 2 .4 60 66.6× ∧ − × + ENTER

f(x) = 0.0875 − 0.4x + 66.6,x2

f(x) = 2x g(x) = 2x + 4 −2 to 2,








Page 414

dFIGURE 7.9

x (x, y) or (x, )

0 (0, 0)1 (1, 2)

x (x, y) or (x, )

0 (0, 4)1 (1, 6)

Check Point 6Graph the functions and in the same rectangular coordinate system. Select integers for x from inclusive. How is the graph ofg related to the graph of f?


f(x) = 2x f(x)

−2 f(−2) = 2(−2) = −4 (−2, −4)

−1 f(−1) = 2(−1) = −2 (−1, −2)

f(0) = 2 ⋅ 0 = 0

f(1) = 2 ⋅ 1 = 2

2

▲

Choose x.

f(2) = 2 ⋅ 2 = 4

▲

Compute f(x) by evaluating f at x

(2, 4)

▲

Form the ordered pair.

g(x) = 2x + 4 g(x)

−2 g(−2) = 2(−2) + 4 = 0 (−2, 0)

−1 g(−1) = 2(−1) + 4 = 2 (−1, 2)

g(0) = 2 ⋅ 0 + 4 = 4

g(1) = 2 ⋅ 1 + 4 = 6

2

▲

Choose x.

g(2) = 2 ⋅ 2 + 4 = 8

▲

Compute g(x) by evaluating g at x

(2, 8)

▲

Form the ordered pair.

f(x) = 2x g(x) = 2x − 3 −2 to 2,








7 Algebra: Graphs, Functions and Linear Systems > 7.1 Graphing and Functions > The Vertical Line Test

The Vertical Line Test

5 Use the vertical line test.

Not every graph in the rectangular coordinate system is the graph of a function. The definition of a function specifies that no value of x can be paired with two or moredifferent values of y. Consequently, if a graph contains two or more different points with the same first coordinate, the graph cannot represent a function. This isillustrated in Figure 7.10. Observe that points sharing a common first coordinate are vertically above or below each other.

FIGURE 7.10 y is not a function of x because 0 is paired with three values of y, namely,

This observation is the basis of a useful test for determining whether a graph defines y as a function of x. The test is called the vertical line test.

TechnologyA graphing calculator is a powerful tool that quickly generates the graph of an equation in two variables. Here is the graph of that we drew by hand inFigure 7.4 on page 410.

What differences do you notice between this graph and the graph we drew by hand? This graph seems a bit “jittery.” Arrows do not appear on the left and right ends ofthe graph. Furthermore, numbers are not given along the axes. For the graph shown above, the x-axis extends from and the y-axis also extends from

The distance represented by each consecutive tick mark is one unit. We say that the viewing window is by

To graph an equation in x and y using a graphing calculator, enter the equation, which must be solved for y, and specify the size of the viewing window. The size of theviewing window sets minimum and maximum values for both the x- and y-axes. Enter these values, as well as the values between consecutive tick marks, on therespective axes. The by viewing rectangle used above is called the standard viewing window.

The Vertical Line Test for FunctionsIf any vertical line intersects a graph in more than one point, the graph does not define y as a function of x.

Example 7 Using the Vertical Line TestUse the vertical line test to identify graphs in which y is a function of x.

a.



1, 0, and − 1.

y = 4 − x2

−10 to 10−10 to 10. [−10, 10, 1] [−10, 10, 1] .

[−10, 10, 1] [−10, 10, 1]









b.

c.

d.

SOLUTION

y is a function of x for the graphs in (b) and (c).

a.

y is not a function of x. Two values of y correspond to one x-value.

b.

y is a function of x.

c.

y is a function of x.

d.



Page 415

y is not a function of x. Two values of y correspond to one x-value.

Check Point 7Use the vertical line test to identify graphs in which y is a function of x.

a.

b.

c.









7 Algebra: Graphs, Functions and Linear Systems > 7.1 Graphing and Functions > Obtaining Information from Graphs

Obtaining Information from Graphs

6 Obtain information about a function from its graph.

Example 8 illustrates how to obtain information about a function from its graph.

Example 8 Analyzing the Graph of a FunctionToo late for that flu shot now! It's only 8 A.M. and you're feeling lousy. Fascinated by the way that algebra models the world (your author is projecting a bit here), youconstruct a graph showing your body temperature from 8 A.M. through 3 P.M. You decide to let x represent the number of hours after 8 A.M. and y represent yourbody temperature at time x. The graph is shown in Figure 7.11. The symbol on the y-axis shows that there is a break in values between 0 and 98. Thus, the firsttick mark on the y-axis represents a temperature of

dFIGURE 7.11 Body temperature from 8 A.M. through 3 P.M.

a. What is your temperature at 8 A.M.?

b. During which period of time is your temperature decreasing?

c. Estimate your minimum temperature during the time period shown. How many hours after 8 A.M. does this occur? At what time does this occur?

d. During which period of time is your temperature increasing?

e. Part of the graph is shown as a horizontal line segment. What does this mean about your temperature and when does this occur?

f. Explain why the graph defines y as a function of x.

SOLUTION

a. Because x is the number of hours after 8 A.M., your temperature at 8 A.M. corresponds to Locate 0 on the horizontal axis and look at the point on thegraph above 0. Figure 7.12 shows that your temperature at 8 A.M. is

dFIGURE 7.12

b. Your temperature is decreasing when the graph falls from left to right. This occurs between and also shown in Figure 7.12. Because xrepresents the number of hours after 8 A.M., your temperature is decreasing between 8 A.M. and 11 A.M.

c. Your minimum temperature can be found by locating the lowest point on the graph. This point lies above 3 on the horizontal axis, shown in Figure 7.13. The y-coordinate of this point falls more than midway between 98 and 99, at approximately 98.6. The lowest point on the graph, (3, 98.6), shows that your minimumtemperature, occurs 3 hours after 8 A.M., at 11 A.M.



98°F.

x = 0.101°F.

x = 0 x = 3,

98.6°F,








Page 416

FIGURE 7.13

d. Your temperature is increasing when the graph rises from left to right. This occurs between and shown in Figure 7.13. Because x represents thenumber of hours after 8 A.M., your temperature is increasing between 11 A.M. and 1 P.M.


x = 3 x = 5,








7 Algebra: Graphs, Functions and Linear Systems > 7.1 Graphing and Functions > Concept and Vocabulary Check

e. The horizontal line segment shown in Figure 7.14 indicates that your temperature is neither increasing nor decreasing. Your temperature remains the same, between and Thus, your temperature is at a constant between 1 P.M. and 3 P.M.

dFIGURE 7.14

f. The complete graph of your body temperature from 8 A.M. through 3 P.M. is shown in Figure 7.14. No vertical line can be drawn that intersects this blue graphmore than once. By the vertical line test, the graph defines y as a function of x. In practical terms, this means that your body temperature is a function of time. Eachhour (or fraction of an hour) after 8 A.M., represented by x, yields precisely one body temperature, represented by y.

Check Point 8When a person receives a drug injected into a muscle, the concentration of the drug in the body, measured in milligrams per 100 milliliters, depends on the timeelapsed after the injection, measured in hours. Figure 7.15 shows the graph of drug concentration over time, where x represents hours after the injection and yrepresents the drug concentration at time x.

dFIGURE 7.15

a. During which period of time is the drug concentration increasing?

b. During which period of time is the drug concentration decreasing?

c. What is the drug's maximum concentration and when does this occur?

d. What happens by the end of 13 hours?

e. Explain why the graph defines y as a function of x.

Concept and Vocabulary CheckFill in each blank so that the resulting statement is true.

1. In the rectangular coordinate system, the horizontal number line is called the ________.

2. In the rectangular coordinate system, the vertical number line is called the ________.

3. In the rectangular coordinate system, the point of intersection of the horizontal axis and the vertical axis is called the ________.

4. The axes of the rectangular coordinate system divide the plane into regions, called ________. There are ________ of these regions.

5. The first number in an ordered pair such as (3, 8) is called the ________. The second number in such an ordered pair is called the ________.



100°F, x = 5 x = 7. 100°F








Page 417

6. The ordered pair (1, 3) is a/an ________ of the equation because when 1 is substituted for x and 3 is substituted for y, we obtain a true statement. Wealso say that (1, 3) ________ the equation.

7. If an equation in two variables (x and y) yields precisely one value of ________ for each value of ________, we say that y is a/an ________ of x.

8. If we can find f(6) by replacing ________ with ________.

9. If any vertical line intersects a graph ________, the graph does not define y as a/an ________ of x.


y = 5x − 2

f(x) = 3x + 5,








7 Algebra: Graphs, Functions and Linear Systems > 7.1 Graphing and Functions > Exercise Set 7.1

Exercise Set 7.1Practice ExercisesIn Exercises 1–20, plot the given point in a rectangular coordinate system.

1. (1, 4)

2. (2, 5)

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13. (0, 0)

14.

15.

16.

17. (3.5, 4.5)

18. (2.5, 3.5)

19.

20.

Graph each equation in Exercises 21–32. Select integers for x from inclusive.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

In Exercises 33–46, evaluate each function at the given value of the variable.

33.

a.



(−2, 3)

(−1, 4)

(−3, −5)

(−4, −2)

(4, −1)

(3, −2)

(−4, 0)

(−5, 0)

(0, −3)

(0, −4)

(−3, −1 )12

(−2, −3 )12

(−5, −2.5)

(1.25, −3.25)

(2.25, −4.25)

−3 to 3,

y = − 2x2

y = + 2x2

y = x − 2

y = x + 2

y = 2x + 1

y = 2x − 4

y = − x12

y = − x + 212

y = x3

y = − 1x3

y = |x| + 1

y = |x| − 1

f(x) = x − 4

f(8)








b.

34.

a.

b.

35.

a.

b.

36.

a.

b.

37.

a.

b.

38.

a.

b.

39.

a.

b.

40.

a.

b.

41.

a.

b.

42.

a.

b.

43.

a.

b.

44.

a.

b.

45.

a.

b.

46.

a.

b.

In Exercises 47–54, evaluate for the given values of x. Then use the ordered pairs from your table to graph the function.

47.

x

f(1)

f(x) = x − 6

f(9)

f(2)

f(x) = 3x − 2

f(7)

f(0)

f(x) = 4x − 3

f(7)

f(0)

g(x) = + 1x2

g(2)

g(−2)

g(x) = + 4x2

g(3)

g(−3)

g(x) = − + 2x2

g(4)

g(−3)

g(x) = − + 1x2

g(5)

g(−4)

h(r) = 3 + 5r2

h(4)

h(−1)

h(r) = 2 − 4r2

h(5)

h(−1)

f(x) = 2 + 3x − 1x2

f(3)

f(−4)

f(x) = 3 + 4x − 2x2

f(2)

f(−1)

f(x) = x

|x|

f(6)

f(−6)

f(x) =|x|

x

f(5)

f(−5)

f(x) (x, f(x))

f(x) = − 1x2

f(x) = − 1x2

−2



x

0 1 2

48.

x

0 1 2

49.

x

0 1 2

50.

x

0 1 2

51.

x

0 1 2 3 4

52.

x

0 1

53.

x

0 1

54.

x

0 1

For Exercises 55–62, use the vertical line test to identify graphs in which y is a function of x.

−1

f(x) = + 1x2

f(x) = + 1x2

−2−1

f(x) = x − 1

f(x) = x − 1

−2−1

f(x) = x + 1

f(x) = x + 1

−2−1

f(x) = (x − 2)2

f(x) = (x − 2)2

f(x) = (x + 1)2

f(x) = (x + 1)2

−3−2−1

f(x) = + 1x3

f(x) = + 1x3

−3−2−1

f(x) = (x + 1)3

f(x) = (x + 1)3

−3−2−1



Page 418

55.

56.

57.

58.

59.

60.









7 Algebra: Graphs, Functions and Linear Systems > 7.1 Graphing and Functions > Exercise Set 7.1

61.

62.

Practice PlusIn Exercises 63–64, let and

63. Find and

64. Find and

In Exercises 65–66, let and g be defined by the following table:

x6 03 4

0 112 0

65. Find

66. Find

In Exercises 67–70, write each English sentence as an equation in two variables. Then graph the equation.

67. The y-value is four more than twice the x-value.

68. The y-value is the difference between four and twice the x-value.

69. The y-value is three decreased by the square of the x-value.

70. The y-value is two more than the square of the x-value.

Application ExercisesA football is thrown by a quarterback to a receiver. The points in the figure show the height of the football, in feet, above the ground in terms of its distance, in yards,from the quarterback. Use this information to solve Exercises 71–76.

d

71. Find the coordinates of point A. Then interpret the coordinates in terms of the information given.



f(x) = − x + 4x2 g(x) = 3x − 5.

g(1) f(g(1)).

g(−1) f(g(−1)).

f(x) g(x)

−2−1

−1−4 −3

−6

− + f(−2) ÷ g(2) ⋅ g(−1).f(−1) − f(0)− −−−−−−−−−√ [g(2)]2

f(1) − f(0) − + g(1) ÷ f(−1) ⋅ g(2) .∣∣ ∣∣ [g(1)]2








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72. Find the coordinates of point B. Then interpret the coordinates in terms of the information given.

73. Estimate the coordinates of point C.

74. Estimate the coordinates of point D.

75. What is the football's maximum height? What is its distance from the quarterback when it reaches its maximum height?

76. What is the football's height when it is caught by the receiver? What is the receiver's distance from the quarterback when he catches the football?

The wage gap is used to compare the status of women's earnings relative to men's. The wage gap is expressed as a percent and is calculated by dividing the median,or middlemost, annual earnings for women by the median annual earnings for men. The bar graph shows the wage gap for selected years from 1980 through 2010.

d

Source: Bureau of Labor Statistics

d

The function models the wage gap, as a percent, x years after 1980. The graph of function G is shown to the right of the actual data. Usethis information to solve Exercises 77–78.

77.

a. Find and interpret Identify this information as a point on the graph of the function.

b. Does overestimate or underestimate the actual data shown by the bar graph? By how much?

78.

a. Find and interpret Identify this information as a point on the graph of the function.

b. Does overestimate or underestimate the actual data shown by the bar graph? By how much?


G(x) = −0.01 + x + 60x2

G(30).

G(30)

G(10).

G(10)








7 Algebra: Graphs, Functions and Linear Systems > 7.2 Linear Functions and Their Graphs

The function models the number of accidents, per 50 million miles driven as afunction of a driver's age, x, in years, where x includes drivers from ages 16 through 74, inclusive. The graph of f isshown. Use the equation for f to solve Exercises 79–82.

d

79. Find and interpret Identify this information as a point on the graph of f.

80. Find and interpret Identify this information as a point on the graph of f.

81. For what value of x does the graph reach its lowest point? Use the equation for f to find the minimum value of y.Describe the practical significance of this minimum value.

82. Use the graph to identify two different ages for which drivers have the same number of accidents. Use theequation for f to find the number of accidents for drivers at each of these ages.

Writing in Mathematics83. What is the rectangular coordinate system?

84. Explain how to plot a point in the rectangular coordinate system. Give an example with your explanation.

85. Explain why and do not represent the same ordered pair.

86. Explain how to graph an equation in the rectangular coordinate system.

87. What is a function?

88. Explain how the vertical line test is used to determine whether a graph represents a function.

Critical Thinking ExercisesMake Sense? In Exercises 89–92, determine whether each statement makes sense or does not make sense, andexplain your reasoning.

89. My body temperature is a function of the time of day.

90. Using I found by applying the distributive property to

91. I knew how to use point plotting to graph the equation so there was really nothing new to learn whenI used the same technique to graph the function

92. The graph of my function revealed aspects of its behavior that were not obvious by just looking at its equation.

In Exercises 93–96, use the graphs of and g to find each number.

d

93.

94.

95.

96.

Table of Contents



f(x) = 0.4 − 36x + 1000x2 f(x),

f(20).

f(50).

(5, −2) (−2, 5)

f(x) = 3x + 2, f(50) (3x + 2)50.

y = − 1,x2

f(x) = − 1.x2

f(−1) + g(−1)

f(1) + g(1)

f(g(−1))

f(g(1))

Thinking Mathematically, SixthEdition

1 Problem Solving and CriticalThinking

2 Set Theory

3 Logic

4 Number Representation andCalculation

5 Number Theory and the RealNumber System

6 Algebra: Equations andInequalities

7 Algebra: Graphs, Functionsand Linear Systems

7.1 Graphing and Functions

7.2 Linear Functions and TheirGraphs

7.3 Systems of Linear Equationsin Two Variables

7.4 Linear Inequalities in TwoVariables

7.5 Linear Programming

7.6 Modeling Data: Exponential,Logarithmic, and QuadraticFunctions

Chapter Summary, Review, andTest

Chapter 7 Test

8 Personal Finance

9 Measurement

10 Geometry

11 Counting Methods andProbability Theory

12 Statistics

13 Voting and Apportionment

14 Graph Theory

Answers to Selected Exercises

Credits

Subject Index





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Technology Exercise97. Use a graphing calculator to verify the graphs that you drew by hand in Exercises 47–54.

7.2 Linear Functions and Their Graphs

What am I Supposed to Learn?After you have read this section, you should be able to:

1 Use intercepts to graph a linear equation.

2 Calculate slope.

3 Use the slope and y-intercept to graph a line.

4 Graph horizontal or vertical lines.

5 Interpret slope as rate of change.

6 Use slope and y-intercept to model data.

IT'S HARD TO BELIEVE THAT THIS gas-guzzler, with its huge fins and overstated design, was available in 1957 forapproximately $1800. Sadly, its elegance quickly faded, depreciating by $300 per year, often sold for scrap just six yearsafter its glorious emergence from the dealer's showroom.

From these casual observations, we can obtain a mathematical model and its graph. The model is


y = −300x + 1800.

▲ ▲

The car is depreciating by

$300 per year for x years.

The new car

is worth $1800.






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7.1 Graphing and Functions - Moreland MATH · 7 Algebra: Graphs, Functions and Linear Systems > 7.1 Graphing and Functions 7.1 Graphing and Functions What am I Supposed to Learn?