Practice Final Solutions: My version of the question is in blue. As usual, the numbers on your question will vary, so be able
to adjust accordingly.
1. Find the center and radius of the circle given by the equation x2 + y 2 − 2x + 4y − 4 = 0.
You’re using completing the square.
x2 + y 2 − 2x + 4y − 4 =
2
+ y 2 + 4y +
=
x − 2x +
2
x − 2x +1 + y 2 + 4y +4 =
(x − 1)2 + (y + 2)2 =
0
4
4+1+4
9
Given problem.
Gather terms, move constant to other side.
Take half of the linear terms for x, y, square it. Add to both sides.
Factor the two quadratics. This is now in standard form.
Thus, your center is C(1, −2) and your radius is r = 3.
2. Find the domain and range of the function shown above.
You’re really just looking for all the possible x and y values, so look at the graph from left to right to get the domain, and
look from bottom to top to get range. Answers will vary, but your answer should use brackets for the interval notation.
3. Write the expression for the line that passes through the point (18, 22) and that is perpendicular to the line 9x + 7y = 218.
You need to use the point-slope formula. You’re given x1 = 18 and y1 = 22. To get m, you need to find the slope of
9
218
the given line 9x + 7y = 218. You do this by solving for y. You should get y = − x +
. The slope you want is a
7
7
7
7
negative reciprocal. Notice here that the constant term is irrelevant. Thus, m = + and your line is y = (x − 18) + 22.
9
9
4. Let f (x) = x6 − 4. Set up the difference quotient and simplify it to an expression that makes sense when h = 0.
To simplify my work, I’ll use h1 (f (x + h) − f (x)) as my difference quotient.
1
h
1
h
1
h
(f (x + h)−f (x))
6
6
−4 −
−4
x+h
x
6
6
−4− +4
x+h
x
1
h
1
h
Apply the difference quotient formula.
6
6
−
x+h x
1
h
1
h
Cancel out the 4.
6x − 6(x + h)
(x − h)(x)
6x − 6x − 6h
(x − h)(x)
−6h
(x − h)(x)
−6
(x − h)(x)
Distribute the negative.
6x
6(x + h)
−
(x + h)x x(x + h)
1
h
Note the colors, and that the −f (x) needs a parenthesis.
Common denominators.
Combine the fraction.
Distribute negative sign.
Cancel the 6x.
Cancel the h.
5. Three years ago a house was worth $230,000. Now the house is worth $330,000. Assume a linear relationship between
time (measured in years) and value (measured in dollars). Find a formula for the value V (t) at time t, where t = 0 refers
to now. What will be the value of the house three years from now?
To compute v(t), you’ll need two ordered pairs. Using (0, 330, 000) and (−3, 230, 000), apply the slope formula and
100, 000
the point slope formula. Remember that your variable should be t. Here, I’ll get V (t) =
(t + 0) + 330, 000.
3
To answer part b, simply plug in t = 3 into your answer from part a.
6. In the diagram above, a right triangle is formed by the coordinate axes and the line segment passing through the points
(0,r), (a, b) = (3,1), and (w,0). Express r as a function of w and L as a function of w.
r−1
0−1
To answer part a, you want to focus on slopes. Using (3, 1) and (0, r): m =
. Using (3, 1) and (w, 0): m =
.
0−3
w−3
Now set these equal and solve for r.
−1
r−1
=
−3
w−3
−→
(r − 1)(w − 3) = 3
−→
r−1=
3
w−3
−→
r =1+
3
w−3
To answer part b, use the Pythagorean Theorem.
s
L2 = w2 + r2
−→
L=
w2 + 1 +
3
w−3
2
7. Find the vertex and x intercept(s) of the parabola y = −12x2 + 21x + 11.
−b
21
To compute vertex, remember h =
=
. To get k, plug this h value into the formula. Remember that WebAssign
2a
24
does not care if you simplify or not.
To get your x-intercepts, use the quadratic formula. Remember that it is ideal to have a positive a value, so move the
terms to the left before you apply the QF. (I’ll let you finish this one on your own.)
8. A right triangle is drawn as shown in the figure below. The base of the triangle is on the x axis and the hypotenuse extends
from the point (−4, 0) to a point in the first quadrant on the (purple) line y = −4x + 33. Write the area of the triangle
as a function of x, then determine the maximum area enclosed by the triangle.
To get the area of the triangle, the base is x + 4 and the height is y = −4x + 33. Thus, A(x) = 0.5(x + 4)(−4x + 33).
You will need to distribute this out to get A(x) = −2x2 + 8.5x + 66.
17 2401
To get the maximum area, you’ll need to find the vertex. The vertex of this quadratic will be at
,
. The area
8 32
is the y-component of your vertex.
9. The cost C to build automobiles is given by a function of the number x of automobiles built. Suppose the number of
automobiles that can be built in t hours is given by the function x given below. Compute the composition (g ◦ f ) as a
function of t. What does the composite function (g ◦ f ) model?
3t + 1.2
C = g(x) = 2400x2 + 7.2
x = f (t) =
2
2
3t + 1.2
To compute (g ◦ f ), you simply input f into g. So (g ◦ f ) = 2400
+ 7.2.
2
Notice that f inputs t in hours and outputs x = number of automobiles. Then g takes x automobiles and converts it
to cost C. Thus the composition will input time t (input of f ) and convert it to cost C (output of g). Therefore, the
composition is the cost of building automobiles as a function of time.
10. Find an exponential function of the form f (x) = ba−x + c that has the given horizontal asymptote y = 31 and y-intercept
211 and passes through point P (2, 111).
For this problem, a normal exponential function has a horizontal asymptote at the x-axis. If this asymptote is now at
y = 31, that means you moved your graph up by 31, so c = 31.
If you use the y-intercept, you get: 211 = b(a)−0 + 21
−→
211 = b(1) + 21
−→
180 = b.
80
1
−2
−2
Now, if you use P , you get:
111 = 180(a) + 31
−→
80 = 180(a)
−→
= 2.
180
a
p
You cross multiply and solve for a to get ± 180/80 = ± 2/3. Use the plus.
11. Compute the inverse of f (x) = − ln(3 − x) and the domain of f −1 (x).
To compute inverses, remember that you first switch your x and y.
x = − ln(3 − y)
−→
−x = ln(3 − y)
−→
e−x = 3 − y
−→
e−x − 3 = −y
−→
−(e−x − 3) = y
To answer part b, notice that this inverse function does not have the three things we’re looking for: division by zero, even
radicals, and logarithmic terms. Thus, its domain is (−∞, ∞).
12. Find the exact solutions to 6x + 216(6−x ) = 42.
Here, we’ll use u-substitution. Let u = 6x . Our equation then becomes u +
you get u2 + 216 = 42u.
216
u
= 42. If you multiply both sides by u,
Move the terms to one side and you’ll get u2 − 42u + 216 = 0, which factors as (u − 36)(u − 6) = 0. Thus,
u = 36
Back Substitute.
u=6
Back Substitute.
6x = 36 Change to Base 6.
6x = 6
Change to Base 6.
6x = 62
6x = 61 .
x=2
x=1
13. Solve the equation log2 (x + 5) = log2 (x − 5) + log3 (9) + 6log6 (3) .
You want to use properties of logs to solve this equation. First, move the log terms to left side.
log2 (x + 5) − log2 (x − 5) = log3 (9) + 6log6 (3)
Use properties of logs on left side, Rewrite 9 in base 3 on right side.
x+5
x−5
x+5
x−5
=
log3 (32 ) + 6log6 (3)
Use cancellation properties on right side.
=
2+3=5
Change your equation to exponential form.
25
=
x+5
x−5
Cross multiply. Note that 25 = 32.
32x − 160
=
x+5
Solve for x.
log2
log2
x = 165/31
Our answer is approximately 5.3226. Remember that when you’re working with a log equation, you must check to see if
your answer works. In this case, it does.
14. Alice invests $4000 at Bob’s bank and $5000 at Charlie’s bank. Bob compounds interest continuously at a nominal rate
of 8%. Charlie compounds interest continuously at a nominal rate of 7%. In how many years will the two investments be
worth the same amount? When both investments are worth the same amount, how much will each be worth?
You’ll get two equations: A = 4000e0.08t and A = 5000e0.07t . Set them equal to each other.
4000e0.08t = 5000e0.07t
−→
4000e0.08t
5000e0.08t
=
4000e0.07t
4000e0.07t
You then convert to log form and divide. Thus, 0.01t = ln(5/4) and t =
e0.01t =
−→
5
4
ln(5/4)
ln(5/4)
. Convert this to
to comply
0.01
1/100
with the NO DECIMAL rule.
To get part b, plug in your answer to part a into either of the two equations. Thus, A = 4000e0.08 ln(5/4)/0.01 ≈ $23, 841.86.
Plug into both equations if you have time to check that they are the same.
15. Express the area of the shaded region as a function of the angle θ, a radian measurement.
The area of the shaded region is the sector (A = 0.5r2 θ) minus the triangle (A = 0.5(base)(height)). Notice that since
you are on the unit circle, the point where the blue line touches the circle is at (x, y) = (cos(θ), sin(θ)).
The area of the sector is A = 0.5(1)2 θ. The area of the triangle is A = 0.5xy = 0.5 cos(θ) sin(θ). Thus, the shaded
region is 0.5θ − 0.5 cos(θ) sin(θ).
16. Assume θ lies in Q2 and the terminal side of θ is parallel to the line y = −13x + 4. Compute sin(θ) and sec(θ).
Draw your triangle in the second
parallel slopes, m = −13/1.
You’ll get that the base = 1,
√ quadrant. Since we have √
√
height = 13, and hypotenuse is 170. Thus, sin(θ) = +13/ 170 and sec(θ) = − 170/1.
17. Compute the exact values:
5π
3π
(a) sin
(b) cos −
6
2
(c) tan(−5π)
(d) cot
5π
4
(e) sec
19π
6
(f) csc
−4π
3
You need to determine where your angles will be. Some will be in a quadrant, while some will be at an axis.
You then apply the appropriate ratio and sign to finish the problem. The answers are:
1
y
0
1
2
(a)
(b) x component = 0
(c) =
(d)
(e) − √
2
x
−1
1
3
2
(f) √
3
18. Use a regular polygon with 14 sides to estimate the area of a circle of radius 8 inches. Then compute your error.
The figure is misleading. The diagram is for an octagon, but it is asking for polygon with more sides. For my example, I
was asked for a polygon with 18 sides inscribed in a circle of radius 6.
2π
Notice that you’ll have 18 such triangles, so the interior angle θ =
. Using the area formula from Section 5.7, we get
18
1
1
2π
that the area of the polygon is A = 18
· a · b sin(θ) = 18
· 6 · 6 sin
.
2
2
18
2π
.
The error estimate will be the circle minus the polygon. Thus, answer = πr2 − polygon = 36π − 324 sin
18
19. Find the exact solutions of the equation on the interval [0, 2π): 8 sin(x) + 8 cos2 (x) = 8.
sin(x) + cos2 (x) = 1
Divide both sides by 8.
sin(x) + 1 − sin2 (x) = 1
Use Pythagorean Identity to substitute for cos2 (x).
2
sin(x) − sin (x) = 0
Cancel out the 1.
sin(x)(1 − sin(x) = 0
Factor out a sin(x).
Thus, sin(x) = 0 or sin(x) = 1. Use the graph or the unit circle to finish this question. Your answers will be at x = 0, π
(when sin(x) = 0) and x = π/2 (When sin(x) = 1.)
20. What is the domain of f (x) = log7 (csc(x))?
We have a logarithmic function, which means that the domain would be when the inside is positive. Thus, we want to
know when csc(x) is positive, or when its graph is above the x-axis. (See red pieces of the graph below.)
Thus, we want (0, π), (2π, 3π), etc. This eliminates the first three choices because we have infinitely many intervals. This
leaves the last two choices and matches D.
21. Compute sin(2θ), cos(2θ) and tan(2θ) if cos(θ) = 3/5 and 0◦ < θ < 90◦ .
Note that θ is in QI. Draw your reference triangle and label your sides. You should have adj = 3, opp = 4, and hyp = 5.
sin(2θ)
sin(2θ) = 2 sin(θ) cos(θ)
cos(2θ) = cos2 (θ) − sin2 (θ)
tan(2θ) =
cos(2θ)
2 2
4
3
3
4
24/25
= 2
=
−
=
5
5
5
5
−7/25
24
9
16
7
24
=
−
=−
= −
25
25 25
25
7
22. A bowl is in the shape of a hemisphere is filled with water to a depth of h = 5 inches. The radius of the bowl is R inches.
Express θ (a radian measure) as a function of R.
=
To do this problem, you need to use the top right triangle. The height is R − 5 and the hypotenuse is R. Thus,
R−5
R−5
sin(θ) =
−→
θ = arcsin
R
R
58π
30
23. Find the exact value of sin arcsin −
and arcsin sin
.
44
55
See your notes from WebQuiz 11 review. For the first part, the sine and arcsine cancels, giving you just −30/44. For the
second part, you need to place 58π/55 on your x-y plane. Find its reference angle (It will be 3π/55.) then use a negative,
since it is in quadrant III and sine is negative there.
24. A bird flies from its nest 10 km in the direction N60◦ E, where it stops to rest on a tree. It then flies 14 km in the direction
due southeast and lands atop a telephone pole. Place an x-y coordinate system so that the origin is the bird’s nest, the
x-axis points east, and the y-axis points north. At what point is the tree located? How far is the telephone pole from the
nest?
Here’s the relevant image with angles and side lengths drawn in:
√
Use the right triangle to compute the coordinates of the tree. You should get that x = 10 cos(30◦ ) = 5 3 and
y = 10 sin(30◦ ) = 5. To answer part b, apply the Law of Cosines, with a = 10, b = 14, θ = 105◦ , and c = unknown.
c2 = a2 + b2 − 2ab cos(θ)
−→
c=
p
102 + 142 − 2 · 10 · 14 · cos(105◦ )
−→
c ≈ 19.1955 km
25. The figure shows a triangle with base c and base angles 38◦ and 76◦ . Express the height of the triangle as a function of c.
This is a two step problem. First, you want to compute a second side of your triangle using Law of Sines. Label the other
two sides a and b.
Here, you can solve for either a or b. I’ll solve for a (in terms of c).
sin(76◦ )
sin(66◦ )
=
a
c
−→
c sin(76◦ ) = a sin(66◦ )
−→
c sin(76◦ )
=a
sin(66◦ )
Now you want to draw your height. You then need to focus on the right triangle using a as your hypotenuse.
Notice that sin(38◦ ) = opp/hyp. Thus,
sin(38◦ ) =
h
a
−→
a · sin(38◦ ) = h −→
c sin(76◦ )
· sin(38◦ ) = h
sin(66◦ )