Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 26 potx

SOLUTION This is the graph of y = x2 shifted to the right 4 units, shrunk vertically to half the height at each x-value, and then shifted downward by 2 units.. If the graph is the graph

6.3 QUADRATICS AND THEIR GRAPHS Graphing Parabolas

Beginning with the basic parabola y = x2and using the results of Section 3.4 on shifting, stretching, shrinking, and flipping, we can obtain any other parabola in which we are interested

EXAMPLE 6.4 Graph f (x) =12(x− 4)2− 2

SOLUTION This is the graph of y = x2 shifted to the right 4 units, shrunk vertically to half the

height at each x-value, and then shifted downward by 2 units (Use the same order

of operations graphically as you would algebraically).8 In particular, the vertex of y =

x2 is shifted right 4 units, to (4, 0), shrunk vertically (no effect, because zero can’t

be “shrunk”), and then shifted down 2 units, to (4, −2) A graph is shown below

y

x

6 4 2

2 4 6 8 –2

(4, –2)

Figure 6.6
Using analogous lines of reasoning, we can graph any function of the form f (x) = a(x− h)2+ k In fact, by using the algebraic technique of completing the square, any quadratic can be expressed in this form.9

Suppose we are given a quadratic in the form f (x) = a(x − r)(x − s), where a, r, and

sare constants This is also convenient to graph We can immediately read off the zeros of the function; f (x) = 0 when x = r or x = s If a > 0, then the parabola opens upward, and

if a < 0, then the parabola opens downward Only parabolas that intersect the x-axis can

be expressed in this form

EXERCISE 6.1 Find the equation of the family of parabolas with vertex (5, 0) Your answer should involve

one parameter, that is, one unspecified constant

Suppose the quadratic in Example 6.4 was not given in the form f (x) = a(x − h)2+ k; suppose we were given this same function in the form

f (x)=1

2x

2

− 4x + 6

To sketch this by hand there are a couple of options One (which we won’t use here) is to put this equation in the form f (x) = a(x − h)2+ k, a process that involves what is called

8 For additional work on order of operations, refer to Appendix A: Algebra.

9

“completing the square”;10the other (given below) is to identify the vertex and one or two other points and draw We can do this without calculus as follows

Identify the x- and y-intercepts:

The y-intercept, f (0), is straightforward to find: f (0) = 6

Finding the x-intercept(s) is equivalent to the problem of solving11 the quadratic equation 12x2− 4x + 6 = 0

1

2x

2

− 4x + 6 = 0

x2− 8x + 12 = 0 Multiply both sides by 2

(x− 6)(x − 2) = 0 Now it’s easier to factor

x− 6 = 0 or x − 2 = 0

x= 6 or x = 2

The parabola has two x-intercepts, one at x = 2 and one at x = 6 Because a parabola is symmetric about its vertical axis, the x-coordinate of the vertex of the parabola lies midway between the two intercepts, at x = 4 The x-coordinate of the vertex is 4, so the y-coordinate must be

f (4) =12(4)2− 4(4) + 6 = 8 − 16 + 6 = −2

Knowing that the vertex is (4, −2) and that the x-intercepts are 2 and 6, we can sketch the graph Knowing the y-intercept is simply a bonus

Alternatively, we could use our knowledge of calculus to find the vertex The x-coordinate of the vertex is the zero of f
(x) We find the y-coordinate as above; identify one additional point on the graph by using symmetry, and sketch

10 See Appendix A: Algebra for details Briefly:

ax2+ bx + c = a



x2+ba +ac



= a



x + b 2a

 2

− b

2

4a 2 +c a



= a



x + b 2a

 2

−b

2 − 4ac 4a .

11 Read Appendix A: Algebra, Part III, Solving Equations, and Section B: Quadratics, for a discussion of solving quadratic

x

6

–2 (4, –2)

Figure 6.7

Notice that if we were handed the function in the form f (x) =12(x− 4)2− 2 and wanted to find the x-intercepts, we would have to solve a quadratic equation Rather than multiplying out, it is more efficient to take advantage of the perfect square as follows

1

2(x− 4)2= 2 (x− 4)2= 4

x− 4 = ±2

x= 4 ± 2

x= 4 + 2 or x = 4 − 2

x= 6 or x = 2

Finding a Function to Fit a Parabolic Graph

Three noncolinear points determine a parabola We might know

i three arbitrary points,

ii the x-intercepts, r1and r2, and a point, or

iii the vertex (p, q) and a point

Depending upon the information available, we can look for an equation of the form

(i) y = ax2+ bx + c (ii) y = k(x − r1)(x− r2) (iii) y − q = k(x − p)2 respectively From equation (i) and three arbitrary points we can obtain a system of three linear equations that we can solve for the constants a, b, and c We can use the x-intercepts

r1and r2in equation (ii) and use the other known point to solve for k Similarly, if we know the vertex and use it in equation (iii) we can solve for k using the other known point

EXAMPLE 6.5 Find a function to fit the parabola drawn below.

y

x

5

1 –1

–2

Figure 6.8

SOLUTION Our strategy is to find an equation of the form

y= k(x − r)(x − s), where r and s are the x-intercepts of the graph First, let’s nail down the x-intercepts; then

we can adjust k as needed (Recall that k can stretch or shrink the graph vertically, flipping vertically if k is negative None of this action will affect the zeros, so we can nail them down with peace of mind.) The x-intercepts are −2 and 3, so −2 and 3 must be zeros of the function The quadratic must be of the form

y= k(x + 2)(x − 3), where k is a constant We know that when x = 0, y = 5, so

5 = k(0 + 2)(0 − 3)

5 = k(−6)

k=−5

6 . Therefore, y =−56(x+ 2)(x − 3)

EXAMPLE 6.6 Find a function to fit the parabola drawn below

y

x

–2 (–1, – 4)

Figure 6.9

Approach I: Again, we nail down the x-intercept It is x = −2, so the quadratic must be of the form

y= k(x + 2)2.

The (x + 2) term must be squared in order to obtain a quadratic.12Alternatively, (−2, 0)

is the vertex of the parabola, so y − 0 = k(x − (−2))2

When x = −1, y = −4, so

−4 = k(−1 + 2)2

−4 = k

y= −4(x + 2)2

Approach II: This is our standard parabola y = x2shifted left 2 units and stretched and flipped Therefore, it is of the form y = k(x + 2)2 Find k as in Approach I We expect

kto be negative to flip the graph, and indeed it is

P R O B L E M S F O R S E C T I O N 6 3

For Problems 1 through 8, graph the function Label the x- and y-intercepts and the coordinates of the vertex.

1 f (x) = −x2+ 1

2 f (x) = (x + 3)(2x − 6)

3 f (x) = x2+ 5x − 6

4 f (x) = 2(x − 1)2− 4

5 f (x) = | − x2+ 1|

6 f (x) = −3(x + 2)2+ 9

7 f (x) = (3 − x)(x + 1)

8 f (x) = x2+ πx + 1

9 Find possible equations for the following graphs assuming that the first three are parabolas and the last is the graph of some cubic

12

Without squaring we would be left with the equation of a line Also, on either side of x = −2 the sign of y doesn’t change,

so (x + 2) must be raised to an even power This type of analysis will be examined in more detail later.

x

(a)

2

2 – 4

(–1, 2)

y

x

(d)

–2

–2 –1

y

x

(b)

2

–1 –2

–3

y

x

(c)

–3

For Problems 10 through 16, give the set of functions all having the specified char-acteristics Example: the set of nonvertical lines passing through ( 3, 2) Solution:

y− 2 = m(x − 3), so f (x) = m(x − 3) + 2, where m is any constant.

10 The set of nonvertical lines passing through (0, π )

11 The set of lines with slope 2

12 The set of parabolas with x-intercepts at x = 0 and x = 3

13 The set of parabolas with x-intercepts at x = −5 and x = 1

14 The set of parabolas with vertex (2, 3)

15 The set of parabolas with vertex (−1, π)

16 The set of parabolas passing through (0, 3) with a slope of 2 at (0, 3)

Strategize In the example above we used y− y1= m(x − x1) as the equation of our

line, not y = mx + b Similarly, when working with parabolas, different forms of the

equation for a parabola are best suited to different situations.

In Problems 17 through 21, find the equation of the parabola with the specifications given.

17 x-intercepts of 3 and −2; maximum value of 1

18 x-intercepts of π and 3π ; y-intercept of 6

19 x-intercepts of π and 3π ; y-intercept of −2

20 Vertex at (1, 5); y-intercept of 1

21 Vertex at (−2, 3); passing through (1, −1)

22 Sketch the graphs of the following functions Beneath the sketch of the function, sketch the graph of the derivative If the graph is the graph of a quadratic, label the coordinates of the vertex of the corresponding parabola; if the graph has corners, label the coordinates of the corners

(a) f (x) = −(x + 2)2 (b) f (x) = (x − 2)2+ 3 (c) f (x) = (x − 2)(x + 4) (d) f (x) = −2(x − 2)(x + 4) (e) f (x) = |x + 4| + 2 (f ) f (x) = |2x| − 3

23 Find the equation of the parabola through the points (0, 3), (1, 0), and (2, −1)

24 Find the coordinates of the vertex of the parabola passing through the points (2, 0), (−1, 9), and (1, −5)

Decide upon a strategy for doing this problem

As mentioned in Section 6.1, quadratic functions arise naturally when modeling the motion

of a falling body We could be interested in the height of a tennis ball thrown in the air, a cannon ball shot from a cannon, a baseball hit by a bat, a football thrown by a quarterback; the list is endless, but the underlying laws of physics are the same Below we’ll look at the free fall of an urban apple

EXAMPLE 6.7 A fellow in a high-rise apartment building decides to get rid of a bad apple by depositing it

in the open trash bin standing on the street directly below his window He opens the window, leans out, and tosses the apple which lands at street level in the bin

Let us assume the following The apple travels vertically only The only force acting

on the apple after it leaves the fellow’s hand is the force of gravity which gives the apple a downward acceleration of 32 ft/sec2 We will ignore the effect of air resistance The laws of physics dictate that the initial height and initial velocity of the apple completely determine its height as a function of time In this case it turns out that

h(t )= −16t2+ 20t + 130, where h(t) is the height of the apple (in feet) t seconds after it is tossed.13 (a) From what height is the apple tossed?

(b) i What is the apple’s initial velocity?

ii Is the apple dropped, tossed down, or tossed up?

13 In the Exploratory Exercise you constructed such a function on your own when given this information plus some initial

(c) i What is the maximum height of the apple?

ii When does it reach this height?

(d) When does the apple land in the trash can?

(e) What is the velocity of the apple when it lands?

(f ) Sketch h(t)

(g) Sketch the graph of velocity versus time

(h) How is the apple’s acceleration reflected in your answer to part (g)?

SOLUTION First go through the questions and determine what is being asked from a mathematical

viewpoint Rewrite each question in terms of the function h(t) Try this on your own and compare your answers with those below

The basic ideas upon which the problem rests are the following:

h(t )gives the height of the apple at time t The graph of h(t) is a parabola opening downward

h
(t )gives the velocity (instantaneous rate of change of height with respect to time) of the apple at time t The graph of h
(t )is a line

|h
(t )| gives the speed of the apple at time t

The derivative of h
(t ), written h

(t ), gives the acceleration (instantaneous rate of change of velocity with respect to time) at time t

Translating the questions into mathematical language gives us the following

(a) What is h(0)?

(b) i What is h
(0)?

ii What is the sign of h
(0)?

If h
(0) is negative, the apple is thrown down

If h
(0) is zero, the apple is dropped

If h
(0) is positive, the apple is tossed up

(c) If the apple is tossed up, then these questions translate to questions about the coordinates

of the vertex

i What is the h-coordinate of the vertex of the parabola?

ii What is the t-coordinate of the vertex?

If the apple is thrown down or dropped, then part (i) is simply the initial height, h(0), and part (ii) is t = 0

(d) Find t such that h(t) = 0 (We want t to be positive.) (e) Let’s denote by tLthe answer to part (d) Evaluate h
(tL)

(f ) Graph the parabola h(t) = −16t2+ 20t + 130 (Be careful to use the appropriate domain.)

(g) Graph h
(t )and interpret it (The domain should be as in part f.) (h) Find h

(t ) This is the slope of the graph of h
(t )

ANSWER (a) h(0) = 130 The apple is tossed from a height of 130 feet

(b) h
(t )= −32t + 20

i h
(0) = 20 The apple’s initial velocity is 20 ft/sec.

ii h
(0) is positive The apple is tossed up

(c) We can find the vertex with or without using calculus Let’s do the former At the vertex h
(t )= 0, the velocity of the apple is zero and the tangent line to the parabola is horizontal Solving the equation h
(t )= −32t + 20 = 0 gives t =−20−32= 5/8

i The h-coordinate of the vertex of the parabola is

h(5/8) = −16 5

8

2 + 20 5 8

 + 130

=−16 · 25

100

4 + 130

=25

4 + 130

= 136.25

Thus, the apple reaches a maximum height of 136.25 feet

ii The t-coordinate of the vertex is 5/8 The apple reaches its maximum height 5/8 seconds after being thrown

(d)

−16t2+ 20t + 130 = 0

−8t2+ 10t + 65 = 0

t=−10 ±

√

100 − 4(−8)(65) 2(−8)

√ 2180

−16

=−10 ± 2

√ 545

−16

=2[−5 ±√545]

−16

√ 545

≈ 3.54 or − 2.29 (numerical approximations)

The answer we are looking for is t =5+√8545, or t ≈ 3.54 seconds, because the apple

hits the trash bin after it is thrown, not before.

(e) h
(t )= −32t + 20



5 +√545 8

= −32



5 +√545 8

+ 20

= −4(5 +√545) + 20

= −20 − 4√545 + 20

= −4√545

≈ −93.38 The apple is falling so the velocity is negative (−4√545 ft/sec or ≈ −93.38 ft/sec), and the speed is positive.14

h

t

5 + √545 8

5

8

Figure 6.10

(f )

t

20 h′ (t) = velocity

5 + √545

8 , – 4 √545

( ( ≈ (3.5, –93.4)

5

8

Figure 6.11

(g) Notice that the velocity is positive when the apple is going up, negative when the apple

is traveling down, and zero at the instant it changes direction (at its maximum height) (h) The acceleration, h

(t ), is −32 ft/sec2, as expected The acceleration is the slope of the velocity graph

14 h
(3.54) = −93.28 The approximation −93.38 is more accurate than this; rounding off after doing all the calculations

produces less error When computing h
(3.54) = −93.28, the error is multiplied by 32.