Polynomial Functions
■ Polynomial Functions
■ Quadratic Functions
■ Graphing Techniques
■ Completing the Square
■ The Vertex Formula
■ Quadratic Models
Polynomial Function
A polynomial function ƒ of degree n, where n is a nonnegative integer, is given by
ƒ1x2 = anxn+ an−1xn−1 + P + a1x +a0, where an, an-1, . . . , a1, and a0 are complex numbers, with an≠0.
In this chapter we primarily consider polynomial functions having real coef- ficients. When analyzing a polynomial function, the degree n and the leading coefficient an are important. These are both given in the dominating term an xn.
LOOKING AHEAD TO CALCULUS In calculus, polynomial functions are used to approximate more complicated functions. For example, the trigono- metric function sin x is approximated by the polynomial
x-x3 6 + x5
120- x7 5040 .
Polynomial Function Function Type Degree n Leading Coefficient an
ƒ1x2 =2 Constant 0 2
ƒ1x2 =5x- 1 Linear 1 5
ƒ1x2 =4x2-x +1 Quadratic 2 4
ƒ1x2 =2x3-12 x+ 5 Cubic 3 2
ƒ1x2 =x4+ 12x3-3x2 Quartic 4 1
Quadratic Function
A function ƒ is a quadratic function if
ƒ1x2 =ax2+ bx+ c, where a, b, and c are complex numbers, with a≠0.
The function ƒ1x2= 0 is the zero polynomial and has no degree.
Quadratic Functions Polynomial functions of degree 2 are quadratic functions. Again, we are most often concerned with real coefficients.
The simplest quadratic function is ƒ1x2 =x2. Squaring function See Figure 1. This graph is a parabola.
Every quadratic function with real coef- ficients defined over the real numbers has a graph that is a parabola. The domain of ƒ1x2 =x2 is 1-∞, ∞2, and the range is 30, ∞2. The lowest point on the graph occurs at the origin 10, 02. Thus, the function decreases on the open interval
y
x (–2, 4)
(–1, 1) (1, 1)
(0, 0) Domain (–∞, ∞)
Range [0, ∞)
f(x) = x2 (2, 4)
Figure 1 x ƒ1x2
-2 4 -1 1
0 0
1 1
2 4
1-∞, 02 and increases on the open interval 10, ∞2. (Remember that these inter- vals indicate x-values.)
331
3.1 Quadratic Functions and Models
Parabolas are symmetric with respect to a line (the y-axis in Figure 1). This line is the axis of symmetry, or axis, of the parabola. The point where the axis intersects the parabola is the vertex of the parabola. As Figure 2 shows, the ver- tex of a parabola that opens down is the highest point of the graph, and the vertex of a parabola that opens up is the lowest point of the graph.
Opens up
Axis
Axis
Opens down Vertex
Vertex
Figure 2
Graphing Techniques Graphing techniques may be applied to the graph of ƒ1x2 = x2 to give the graph of a different quadratic function. Compared to the basic graph of ƒ1x2= x2, the graph of F1x2 = a1x- h22 + k, with a≠0, has the following characteristics.
F1x2 =a1x −h22 +k
• Opens up if a70
• Opens down if a60
• Vertically stretched (narrower) if a71
• Vertically shrunk (wider) if 06a61
Horizontal shift:
• h units right if h70
• h units left if h60
Vertical shift:
• k units up if k70
• k units down if k60
EXAMPLE 1 Graphing Quadratic Functions Graph each function. Give the domain and range.
(a) ƒ1x2= x2 - 4x- 2 (by plotting points)
(b) g1x2= -12 x21and compare to y= x2 and y = 12 x22
(c) F1x2 = -121x -422 + 3 (and compare to the graph in part (b)) SOLUTION
(a) See the table with Figure 3. The domain of ƒ1x2 =x2 - 4x- 2 is 1-∞, ∞2, the range is 3-6, ∞2, the vertex is the point 12, -62, and the axis has equa- tion x= 2. Figure 4 shows how a graphing calculator displays this graph.
(b) Think of g1x2= -12 x2 as g1x2= -A12 x2B. The graph of y = 12 x2 is a wider version of the graph of y= x2, and the graph of g1x2= -A12 x2B is a reflec- tion of the graph of y =12 x2 across the x-axis. See Figure 5 on the next page.
The vertex is the point 10, 02, and the axis of the parabola is the line x =0 (the y-axis). The domain is 1-∞, ∞2, and the range is 1-∞, 04.
2 x = 2
–6 –2 3
x y
0
f(x) = x2 – 4x – 2
(2, –6) Figure 3 x ƒ1x2
-1 3 0 -2 1 -5 2 -6 3 -5 4 -2
5 3
Figure 4 f(x) = x2 − 4x−2
−10
−10
10
10
Calculator graphs are shown in Figure 6.
(c) Notice that F1x2 = -121x- 422 +3 is related to g1x2 = -12 x2 from part (b).
The graph of F1x2 is the graph of g1x2 translated 4 units to the right and 3 units up. See Figure 7. The vertex is the point 14, 32, which is also shown in the calculator graph in Figure 8, and the axis of the parabola is the line x =4. The domain is 1-∞, ∞2, and the range is 1-∞, 34.
✔ Now Try Exercises 19 and 21.
Completing the Square In general, the graph of the quadratic function ƒ1x2 = a1x− h22 + k 1a 3 02
is a parabola with vertex 1h, k2 and axis of symmetry x= h. The parabola opens up if a is positive and down if a is negative. With these facts in mind, we complete the square to graph the general quadratic function
ƒ1x2 =ax2+ bx+ c.
EXAMPLE 2 Graphing a Parabola 1a=12
Graph ƒ1x2= x2 - 6x+ 7. Find the largest open intervals over which the func- tion is increasing or decreasing.
SOLUTION We express x2 - 6x+ 7 in the form 1x- h22 + k by completing the square. In preparation for this, we first write
ƒ1x2 =1x2 - 6x 2 + 7. Prepare to complete the square.
We must add a number inside the parentheses to obtain a perfect square trino- mial. Find this number by taking half the coefficient of x and squaring the result.
y
x y = x2
(0, 0) x = 0
g(x) = – x12 2 y = x12 2
Figure 5
−5
−8
5
8
g(x) = – x1 2 2
y = x1 2 y = x2 2
Figure 6
x y
(0, 0)
4
F(x) = – (x – 4)2 + 3
x = 4 (4, 3)
1 2
g(x) = – x12 2
Figure 7
−6
−3
4
8
g(x) = – x1 2 2
F(x) = − 1 (x − 4)2+ 3 2
Figure 8
333
3.1 Quadratic Functions and Models
C121-62D2 =1-322 =9 Take half the coefficient of x.
Square the result.
ƒ1x2 =1x2 - 6x+ 9- 92 +7 Add and subtract 9.
ƒ1x2 =1x2 - 6x+ 92- 9 +7 Regroup terms.
ƒ1x2 =1x - 322- 2 Factor and simplify.
The vertex of the parabola is the point 13, -22, and the axis is the line x= 3. We find additional ordered pairs that satisfy the equation, as shown in the table, and plot and join these points to obtain the graph in Figure 9.
This is the same as adding 0.
f(x) = x2 − 6x + 7 f(x) = (x − 3)2 − 2
−4
−3
10
9
This screen shows that the vertex of the graph in Figure 9 is the point 13, -22. Because it is the lowest point on the graph, we direct the calculator to find the minimum.
x
f(x) = x2 – 6x + 7 f(x) = (x – 3)2 – 2 y
0 3
7
6 –2 (3, –2)
x = 3
Figure 9
The domain of this function is 1-∞, ∞2, and the range is 3-2, ∞2. Because the lowest point on the graph is the vertex 13, -22, the function is decreasing on 1-∞, 32 and increasing on 13, ∞2.
✔ Now Try Exercise 31.
NOTE In Example 2 we added and subtracted 9 on the same side of the equation to complete the square. This differs from adding the same number to each side of the equation, as is sometimes done in the procedure. We want ƒ1x2:that is, y—alone on one side of the equation, so we adjusted that step in the process of completing the square here.
EXAMPLE 3 Graphing a Parabola 1a312
Graph ƒ1x2= -3x2- 2x+ 1. Identify the intercepts of the graph.
SOLUTION To complete the square, the coefficient of x2 must be 1.
ƒ1x2= -3ax2 + 2
3 x b + 1 Factor -3 from the first two terms.
ƒ1x2= -3ax2 + 2 3 x+ 1
9 - 1
9b + 1 C12A23BD2=A13B2=19 , so add and subtract 19 .
ƒ1x2= -3ax2 + 2 3 x+ 1
9b - 3a-1
9b + 1 Distributive property ƒ1x2= -3ax + 1
3b2 + 4
3 Factor and simplify.
The vertex is the point A-13 , 43B. The intercepts are good additional points to find.
The y-intercept is found by evaluating ƒ102.
ƒ102 = -31022- 2102 +1 Let x=0 in ƒ1x2= -3x2-2x+1.
ƒ102 =1 The y-intercept is 10, 12.
Be careful here.
x y
0 7
1 2
3 -2
5 2
6 7
Find using symmetry about the axis.
y-intercept Vertex
This screen gives the vertex of the graph in Figure 10 as the point
A-13 , 43B. (The display shows decimal approximations.) We want the highest point on the graph, so we direct the calculator to find the maximum.
f(x) = −3x2 − 2x + 1 f(x) = −3(x + )2 +
−4.1
−6.6
4.1
6.6
1 3
4 3
The x-intercepts are found by setting ƒ1x2 equal to 0 and solving for x.
0 = -3x2 - 2x+ 1 Set ƒ1x2=0.
0 = 3x2+ 2x- 1 Multiply by -1.
0 = 13x- 121x+ 12 Factor.
x = 1
3 or x = -1 Zero-factor property
Therefore, the x-intercepts are A13, 0B and 1-1, 02. The graph is shown in Figure 10.
x 2 1 –1
( )– ,
y 13 4
3
f(x) = –3x2 – 2x + 1 f(x) = –3(x + )132 +4
3 0
2 1
x = –13 1 3 (0, 1) (–1, 0)
( , 0)
Figure 10 ✔ Now Try Exercise 33.
NOTE It is possible to reverse the process of Example 3 and write the quadratic function from its graph in Figure 10 if the vertex and any other point on the graph are known. Because quadratic functions take the form
ƒ1x2= a1x- h22 +k,
we can substitute the x- and y-values of the vertex, A-13 , 43B, for h and k.
ƒ1x2= acx - a-1
3b d2 + 4
3 Let h= -
1
3 and k=43 . ƒ1x2= aax+ 1
3b2 + 4
3 Simplify.
We find the value of a by substituting the x- and y-coordinates of any other point on the graph, say 10, 12, into this function and solving for a.
1= aa0+ 1 3b2 + 4
3 Let x=0 and y=1.
1= aa1 9b + 4
3 Square.
-1 3 = 1
9 a Subtract 43 .
a= -3 Multiply by 9. Interchange sides.
Verify in Example 3 that the vertex form of the quadratic function is ƒ1x2= -3 ax+ 1
3b2+ 4 3 .
Exercises of this type are labeled Connecting Graphs with Equations.
335
3.1 Quadratic Functions and Models
The Vertex Formula We can generalize the earlier work to obtain a for- mula for the vertex of a parabola.
ƒ1x2= ax2 +bx + c General quadratic form
= aax2 + b
a x b + c Factor a from the first two terms.
= aax2 + b a x+ b2
4a2b + c - aa b2 4a2b
Add C12AbaBD2=4ab22 inside the parentheses. Subtract aA4ab22B
outside the parentheses.
= aax+ b
2ab2 + c- b2
4a Factor and simplify.
ƒ1x2= aJx- a- b
2abR2 + 4ac -b2
4a Vertex form of ƒ1x2=a1x-h22+k (11)11* (111)111*
h k
Thus, the vertex 1h, k2 can be expressed in terms of a, b, and c. It is not neces- sary to memorize the expression for k because it is equal to ƒ1h2 = ƒA−2ab B.
LOOKING AHEAD TO CALCULUS An important concept in calculus is the definite integral. If the graph of ƒ lies above the x-axis, the symbol
L
b a
ƒ1x2 dx
represents the area of the region above the x-axis and below the graph of ƒ from x=a to x=b. For example, in
Figure 10 with
ƒ1x2= -3x2-2x+1, a= -1, and b=13 , calculus provides the tools for determining that the area enclosed by the parabola and the x-axis is 3227 (square units).
Graph of a Quadratic Function
The quadratic function ƒ1x2 = ax2+ bx +c can be written as y= ƒ1x2 = a1x− h22+ k, with a≠ 0,
where h= − b
2a and k =ƒ1h2. Vertex formula The graph of ƒ has the following characteristics.
1. It is a parabola with vertex 1h, k2 and the vertical line x =h as axis.
2. It opens up if a 70 and down if a 6 0.
3. It is wider than the graph of y= x2 if a 61 and narrower if a 7 1.
4. The y-intercept is 10, ƒ1022 = 10, c2.
5. The x-intercepts are found by solving the equation ax2+ bx +c = 0.
• If b2- 4ac 70, then the x-intercepts are Q-b{ 22ab2- 4ac, 0R .
• If b2 -4ac =0, then the x-intercept is A-2ab , 0B.
• If b2 -4ac 60, then there are no x-intercepts.
EXAMPLE 4 Using the Vertex Formula
Find the axis and vertex of the parabola having equation ƒ1x2 = 2x2 + 4x+ 5.
SOLUTION The axis of the parabola is the vertical line x =h = - b
2a = - 4
2122 = -1. Use the vertex formula.
Here a=2 and b=4.
The vertex is 1-1, ƒ1-122. Evaluate ƒ1-12.
ƒ1-12 = 21-122 +41-12+ 5= 3
The vertex is 1-1, 32. ✔ Now Try Exercise 31(a).
Quadratic Models Because the vertex of a vertical parabola is the highest or lowest point on the graph, equations of the form
y= ax2 + bx+ c
are important in certain problems where we must find the maximum or mini- mum value of some quantity.
• When a60, the y-coordinate of the vertex gives the maximum value of y.
• When a70, the y-coordinate of the vertex gives the minimum value of y.
The x-coordinate of the vertex tells where the maximum or minimum value occurs.
If air resistance is neglected, the height s (in feet) of an object projected directly upward from an initial height s0 feet with initial velocity v0 feet per second is
s1t2 = −16 t2 +v0 t +s0 ,
where t is the number of seconds after the object is projected. The coefficient of t21that is, -162 is a constant based on the gravitational force of Earth. This constant is different on other surfaces, such as the moon and the other planets.
EXAMPLE 5 Solving a Problem Involving Projectile Motion
A ball is projected directly upward from an initial height of 100 ft with an initial velocity of 80 ft per sec.
(a) Give the function that describes the height of the ball in terms of time t.
(b) After how many seconds does the ball reach its maximum height? What is this maximum height?
(c) For what interval of time is the height of the ball greater than 160 ft?
(d) After how many seconds will the ball hit the ground?
ALGEBRAIC SOLUTION
(a) Use the projectile height function.
s1t2= -16t2 +v0t +s0 Let v0=80 and s0=100.
s1t2= -16t2 +80t + 100
(b) The coefficient of t2 is -16, so the graph of the projectile function is a parabola that opens down.
Find the coordinates of the vertex to determine the maximum height and when it occurs. Let a= -16 and b= 80 in the vertex formula.
t = - b
2a = - 80
21-162 = 2.5 s1t2= -16t2 + 80t +100
s12.52= -1612.522 +8012.52+ 100 s12.52= 200
Therefore, after 2.5 sec the ball reaches its maxi- mum height of 200 ft.
GRAPHING CALCULATOR SOLUTION
(a) Use the projectile height function as in the alge- braic solution, with v0= 80 and s0 = 100.
s1t2 = -16t2 + 80t + 100
(b) Using the capabilities of a calculator, we see in
Figure 11 that the vertex coordinates are indeed 12.5, 2002.
Be careful not to misinterpret the graph in Fig- ure 11. It does not show the path followed by the ball. It defines height as a function of time.
y1 = −16x2 + 80x + 100
0 300
−70
9.4
Figure 11
Here x=t and y1=s1t2.
337
3.1 Quadratic Functions and Models
(c) If we graph
y1 = -16x2 +80x +100 and y2 = 160, as shown in Figures 12 and 13, and locate the two points of intersection, we find that the x-coordinates for these points are approximately
0.92 and 4.08.
Therefore, between 0.92 sec and 4.08 sec, y1 is greater than y2, and the ball is greater than 160 ft above the ground.
(c) We must solve the related quadratic inequality.
-16t2 + 80t+ 1007160
-16t2 + 80t- 6070 Subtract 160.
4t2 - 20t+ 1560 Divide by -4. Reverse the inequality symbol.
Use the quadratic formula to find the solutions of 4t2- 20t + 15= 0.
t = -1-202{ 21-2022 - 41421152 2142
Here a=4, b= -20, and c=15.
t = 5- 210
2 ≈0.92 or t= 5+ 210
2 ≈4.08
These numbers divide the number line into three intervals:
1-∞, 0.922, 10.92, 4.082, and 14.08, ∞2. Using a test value from each interval shows that 10.92, 4.082 satisfies the inequality. The ball is greater than 160 ft above the ground between 0.92 sec and 4.08 sec.
(d) The height is 0 when the ball hits the ground.
We use the quadratic formula to find the positive solution of the equation
-16t2 + 80t+ 100= 0.
Here, a= -16, b =80, and c = 100.
t = -80{2802- 41-16211002 21-162
t ≈ -1.04 or t≈6.04 Reject
The ball hits the ground after about 6.04 sec.
(d) Figure 14 shows that the x-intercept of the graph of y = -16x2 +80x + 100 in the given window is approximately 16.04, 02, which means that the ball hits the ground after about 6.04 sec.
✔ Now Try Exercise 57.
EXAMPLE 6 Modeling the Number of Hospital Outpatient Visits The number of hospital outpatient visits (in millions) for selected years is shown in the table.
Source: American Hospital Association.
Year Visits Year Visits
99 573.5 106 690.4
100 592.7 107 693.5
101 612.0 108 710.0
102 640.5 109 742.0
103 648.6 110 750.4
104 662.1 111 754.5
105 673.7 112 778.0
−6 0
−70 300
9.4
y2 = 160 y1 = −16x2 + 80x + 100
Figure 12
−6 0
−70 300
9.4
y2 = 160 y1 = −16x2 + 80x + 100
Figure 13
0
−70 300
9.4
y1 = −16x2 + 80x + 100
Figure 14
In the table on the preceding page, 99 represents 1999, 100 represents 2000, and so on, and the number of outpatient visits is given in millions.
(a) Prepare a scatter diagram, and determine a quadratic model for these data.
(b) Use the model from part (a) to predict the number of visits in 2016.
SOLUTION
(a) Linear regression is used to determine linear equations that model data.
With a graphing calculator, we can use quadratic regression to find qua- dratic equations that model data.
The scatter diagram in Figure 15(a) suggests that a quadratic function with a negative value of a (so the graph opens down) would be a reasonable model for the data. Using quadratic regression, the quadratic function
ƒ1x2= -0.16161x2 +49.071x - 2695.5
approximates the data well. See Figure 15(b). The quadratic regression values of a, b, and c are displayed in Figure 15(c).
(b) The year 2016 corresponds to x = 116. The model predicts that there will be 822 million visits in 2016.
ƒ1x2 = -0.16161x2 + 49.071x- 2695.5
ƒ 11162 = -0.16161111622 + 49.07111162- 2695.5 ƒ11162≈822 million
✔ Now Try Exercise 73.
(a)
99500 800
113
99500 800
113
f(x) = −0.16161x2 + 49.071x − 2695.5
(b)
(c) Figure 15
CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence.
1. A polynomial function with leading term 3x5 has degree .
2. The lowest point on the graph of a parabola that opens up is the of the parabola.
3. The highest point on the graph of a parabola that opens down is the of the parabola.
4. The axis of symmetry of the graph of ƒ1x2= 21x +422- 6 has equation x= . 5. The vertex of the graph of ƒ1x2= x2+2x +4 has x-coordinate .
6. The graph of ƒ1x2= -2x2-6x +5 opens down with y-intercept 10, 2, so it
has x-intercept(s).
(no/one/two)