Applications of Polynomials

CHAPTER 5. POLYNOMIALS
290
Use the Calculator Submission Guidelines when reporting your solution.
1. Draw axes with a
ruler.
p(x) = x4 − 10x3 − 4x2 + 250x − 525
y
500
2. Label the horizontal
axis x and the
vertical axis y.
3. Indicate the
WINDOW
parameters Xmin,
Xmax, Ymin, and
Ymax, at the end of
each axis.
−10
4. Freehand the curve
and label it with its
equation.
5.3
10
x
−1000
Applications of Polynomials
1. To find the firm’s revenue when it spends $10,000 on advertising, locate 10
on the horizontal axis (the horizontal axis is measured in thousands of dollars),
draw a vertical arrow to the curve, then a horizontal arrow to the vertical axis.
It appears that this last arrow points at a number between 7,400 and 7,500 on
the vertical axis. We’ll guess 7,450. Because the vertical axis is measured in
thousands of dollars, the firm’s revenue is approximately $7,450,000.
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291
R (thousands of dollars)
8,500
8,000
7,500
7,000
6,500
6,000
x (thousands of dollars)
0
5
10
15
20
25
30
To use the polynomial to find the firm’s revenue, substitute 10 for x.
R(x) = −4.1x2 + 166.8x + 6196
R(10) = −4.1(10)2 + 166.8(10) + 6196
Using a calculator, we estimate:
R(1) ≈ 7454
Hence, when the firm invests $10,000 in advertising, their revenue is approximately $7,454,000.
3. To find the concentration of medication in the patient’s blood after 2 hours,
locate the number 2 on the horizontal axis, draw a vertical arrow to the graph,
then a horizontal arrow to the vertical axis. It appears that this last arrow
points at a number slightly above 60 mg/L. We’ll guess 63 mg/L.
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292
C (mg/L)
100
80
60
40
20
t (Hours)
0
0
1
2
3
To use the polynomial to find the medication concentration after 2 hours,
substitute 2 for t in the given polynomial.
C(t) = −56.214t2 + 139.31t + 9.35
C(2) = −56.214(2)2 + 139.31(2) + 9.35
Using a calculator, we estimate:
N (3) ≈ 63
Hence, the concentration of medication in the patient’s blood after 2 hours is
approximately 63 mg/L.
5. The projectile’s height above ground is given by the formula
1
y = y0 + v0 t − gt2 ,
2
where the initial height is 75 meters, the initial velocity is v0 = 457 meters per
second, and the acceleration due to gravity is g = 9.8 meters per second per
second. We’re asked to find when the object first reaches a height of y = 6592
meters. Substituting these numbers, the formula becomes
1
6592 = 75 + 457t − (9.8)t2 ,
2
or equivalently,
6592 = 75 + 457t − 4.9t2 .
Load each side of this equation into the Y= menu of your graphing calculator,
then set the WINDOW parameters as follows: Xmin=0, Xmax=100, Ymin=0,
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and Ymax=10800. Push the GRAPH button to produce the graph, then use
the 5:intersect tool from the CALC menu to find the first time the height of
the object reaches 6592 meters. Report the results on your homework as shown
in the following graph.
y
10,800
y = 75 + 457t − 4.9t2
y = 6592
0
0
17.6
100
t
Hence, the projectile first reaches a height of 6592 meters at approximately
17.6 seconds.
7. The projectile’s height above ground is given by the formula
1
y = y0 + v0 t − gt2 ,
2
where the initial height is 58 meters, the initial velocity is v0 = 229 meters per
second, and the acceleration due to gravity is g = 9.8 meters per second per
second. We’re asked to find when the object first reaches a height of y = 1374
meters. Substituting these numbers, the formula becomes
1
1374 = 58 + 229t − (9.8)t2 ,
2
or equivalently,
1374 = 58 + 229t − 4.9t2 .
Load each side of this equation into the Y= menu of your graphing calculator,
then set the WINDOW parameters as follows: Xmin=0, Xmax=50, Ymin=0,
and Ymax=2800. Push the GRAPH button to produce the graph, then use
the 5:intersect tool from the CALC menu to find the first time the height of
the object reaches 1374 meters. Report the results on your homework as shown
in the following graph.
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CHAPTER 5. POLYNOMIALS
294
y
y = 58 + 229t − 4.9t2
2,800
y = 1374
0
0
6.7
50
t
Hence, the projectile first reaches a height of 1374 meters at approximately 6.7
seconds.
9. To find the zero of f (x) = 3.25x − 4.875 algebraically, set f (x) = 0 and
solve for x.
f (x) = 0
3.25x − 4.875 = 0
We want the value of x that makes
the function equal to zero.
Replace f (x) with 3.25x − 4.875.
Now, solve for x.
3.25x = 4.875
4.875
3.25x
=
3.25
3.25
x = 1.5
Add 4.875 to both sides.
Divide both sides by 3.25.
Simplify.
Hence, 1.5 is a zero of f (x) = 3.25x − 4.875. Next, load the function into the
Y= menu, select 6:ZStandard from the ZOOM menu, then use the utility
2:zero from the CALC menu to find the zero of the function.
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Next, use the Calculator Submission Guidelines when crafting the following
report on your homework paper. Draw a dashed vertical line through the xintercept and label it with its coordinates.
y
f
10
−10
(1.5, 0)
10
x
−10
Note how the graphical solution agrees with the algebraic solution.
11. To find the zero of f (x) = 3.9 − 1.5x algebraically, set f (x) = 0 and solve
for x.
f (x) = 0
3.9 − 1.5x = 0
We want the value of x that makes
the function equal to zero.
Replace f (x) with 3.9 − 1.5x.
Now, solve for x.
−1.5x = −3.9
−1.5x
−3.9
=
−1.5
−1.5
x = 2.6
Subtract 3.9 from both sides.
Divide both sides by −1.5.
Simplify.
Hence, 2.6 is a zero of f (x) = 3.9 − 1.5x. Next, load the function into the Y=
menu, select 6:ZStandard from the ZOOM menu, then use the utility 2:zero
from the CALC menu to find the zero of the function.
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CHAPTER 5. POLYNOMIALS
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Next, use the Calculator Submission Guidelines when crafting the following
report on your homework paper. Draw a dashed vertical line through the xintercept and label it with its coordinates.
y
f
10
(2.6, 0)
−10
10
x
−10
Note how the graphical solution agrees with the algebraic solution.
13. The projectile’s height above ground is given by the formula
1
y = y0 + v0 t − gt2 ,
2
where the initial height is 52 meters, the initial velocity is v0 = 203 meters per
second, and the acceleration due to gravity is g = 9.8 meters per second per
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second. Substituting these numbers, the formula becomes
1
y = 52 + 203t − (9.8)t2 ,
2
or equivalently,
y = 52 + 203t − 4.9t2 .
We’re asked to find the time it takes the projectile to return to ground level.
When this happens, its height y above ground level will equal zero. Substitute
0 for y in the last equation.
0 = 52 + 203t − 4.9t2 .
Load the right-hand side of this equation into the Y= menu of your graphing
calculator, then set the WINDOW parameters as follows: Xmin=0, Xmax=50,
Ymin=0, and Ymax=2200. Push the GRAPH button to produce the graph,
then use the 2:zero tool from the CALC menu to find where the graph of
f crosses the x-axis. Report the results on your homework as shown in the
following graph.
y(meters)
y = 52 + 203t − 4.9t2
2,200
0
0
50
41.7
t(seconds)
Hence, the projectile reaches ground level at approximately 41.7 seconds.
15. The projectile’s height above ground is given by the formula
1
y = y0 + v0 t − gt2 ,
2
where the initial height is 52 meters, the initial velocity is v0 = 276 meters per
second, and the acceleration due to gravity is g = 9.8 meters per second per
second. Substituting these numbers, the formula becomes
1
y = 52 + 276t − (9.8)t2 ,
2
or equivalently,
y = 52 + 276t − 4.9t2 .
Second Edition: 2012-2013