INTERNET MAT 117
Solution for the Review Problems
(1) Let us consider the circle with equation
3
= 0.
4
(a) Find the standard form of the equation of the circle given above.
(i) Group the x and y terms together and take the constant term
to the other side.
3
x2 + 2x + y 2 + 3y = −
4
(ii) Complete the square by adding the square of half of the coefficients of x and y to both sides of the equation. Add 1 and
32
= 94 to both sides of the equation.
22
x2 + y 2 + 2x + 3y +
9
9 3
=1+ −
4
4 4
(iii) Complete the square, combine the constants. See the standard form below:
x2 + 2x + 1 + y 2 + 3y +
5
3 2 10
=
(x + 1)2 + (y + ) =
2
4
2
(b) Find the center and the radius of the circle given above.
3
Center: (−1, − ).
2
p
Radius: 5/2 = 1.58113883.
(2) (a) Find the slope intercept form of the line passing through (−2, 5)
and parallel to the line 5x − 3y = 8.
(i) Find the slope intercept form of the line 5x − 3y = 8 by
expressing y in terms of x.
−3y = −5x + 8
5
8
y = x−
3
3
5
Thus, the slope of the line 5x − 3y = 8 is . Hence, the slope
3
5
of a parallel line to the line 5x − 3y = 8 is also .
3
2
(ii) Now we have to find the equation of the line passing through
5
the point (−2, 5) that has the slope . The slope intercept
3
5
form of any line with slope can be written in the form of
3
5
y = x + b.
3
The line passing through the point (−2, 5). Substitute x = −2
and y = 5 and solve the equation for b.
5=−
10
+ b.
3
25
3
So the equation of the line passing through the point (−2, 5)
and has the slope 35 is
b=
25
5
y = x+ .
3
3
(b) Find the slope intercept form of the line passing through (−2, 5)
and perpendicular to the line 5x − 3y = 8.
(i) Find the slope intercept form of the line 5x − 3y = 8 by
expressing y in terms of x
−3y = −5x + 8
5
8
y = x−
3
3
5
Thus, the slope of the line 5x − 3y = 8 is . Hence, the slope
3
3
of a perpendicular line to the line 5x − 3y = 8 is − .
5
(ii) Now we have to find the equation of the line passing through
3
the point (−2, 5) that has the slope − . The slope intercept
5
3
form of any line with slope − can be written in the form of
5
3
y = − x + b.
5
The line passing through the point (−2, 5). Substitute x = −2
and y = 5 and solve the equation for b.
5=
6
+ b.
5
b=
19
5
3
So, the equation of the line passing through the point (−2, 5)
3
and that the slope − is
5
19
3
y =− x+ .
5
5
(3) Find the domain of the following functions. Give your answer in interval
notation.
√
x−1
(a) g(x) =
5−x
(i) The denominator cannot be zero , so x 6= 5.
(ii) The expression under the radical cannot be negative, so 1 ≤ x.
Hence 1 ≤ x and x 6= 5. With interval notation,
[1, 5) ∪ (5, ∞).
x+1
(b) f (x) =
4 − x2
(i) The denominator cannot be zero , so x2 − 4 6= 0 that is x 6= 2
and x 6= −2. With interval notation
(∞, −2) ∪ (−2, 2) ∪ (2, ∞).
(4) A small business buys a computer for $ 4,000. After 4 years the value
of the computer is expected to be $ 200. For accounting purposes, the
business uses linear depreciation to assess the value of the computer at
a given time. This means the if V is the value of the computer at the
time t, then a linear equation is used to relate V and t.
(a) Find a linear equation that relates V and t.
Slope of the line passing through the points (0, 4000) and (4,200)
is −950. The equation of the line passing through the give points
with slope −950 is
V (t) = −950t + 4000
(b) Find the depreciated values of the computer 3 years from the date
of purchase.
Substitute t = 3 years in to the equation obtained in part a. Then
V = 1150. So, it is $1,150.
√
(5) Let f (x) = x2 − 1 and g(x) = x + 2
f
f
(a) Find ( )(x) and give the domain of .
g
g
x2 − 1
f
.
( )(x) = √
g
x+2
4
(i) The denominator can not be zero and the expression can not
be negative under the radical, so x > −2.
f
(ii) The domain of ( )(x) is (−2, ∞) since x > −2.
g
(b) Find (f ◦ g)(x) and (g ◦ f )(x)
(f ◦ g)(x) = x
√+ 2 − 1 = x +√1
(g ◦ f )(x) = x2 − 1 + 2 = x2 + 1
(6) Let f (x) = −2x + 4.
(a) Find f −1 (x).
y = −2x + 4
(i) Solve the equation for x. (Express x in terms of y)
y − 4 = −2x
4−y
y−4
= x or
=x
−2
2
(ii) Interchange the variables x and y.
4−x
=y
2
4−x
f −1 (x) =
2
(b) Graph f and f −1 on the same coordinate axis.
6
4
y = −2x+4
y = (4−x)/2
2
0
−2
−4
−6
−6
−4
−2
0
2
4
6
5
(7) Use the graph of f (x) below to sketch the graph of g(x) it if is known
that g(x) = f (x − 1) + 20. Draw g(x) on the same set of axes as the
given graph.
80
f(x−1)+20
60
f(x−1)
f(x)
40
20
0
−20
−40
−4
−3
−2
−1
0
1
2
3
4
5
6
7
6
(8) Consider the graph of the function f (x) below. Answer the following
questions:
5
4
3
2
1
0
−1
−2
−3
−4
−3
−2
−1
0
1
2
3
4
(a) Use interval notation to state the domain of f (x).
[−2, 4]
(b) Use interval notation to state the range of f (x).
[−1, 4]
(c) State the interval where the function f (x) increases.
[0, 1)
(d) State the interval where the function f (x) decreases.
[−2, 0] ∪ [1, 4]
(e) State the values of f (1) and f (4).
f (1) = 2 f (4) = −1
5
6
7
(9) Let P (x) = x3 − 3x2 − 9x − 5
(a) Determine all the possible rational zeros for P (x).
By the Rational Zeros Theorem if pq is a zero of P (x) then p divides
5 and q divides 1, so pq in the form of
factor of 5
.
factor of 1
The factors of 5 are 1,-1, 5, -5 and the factors of 1 are 1 and -1.
Thus, the possible rational zeros pq are 1, -1, 5, -5.
(b) Factor the polynomial P (x) completely using the procedure of long
division or synthetic division.
5 is a zero, that is P(5)=0. Using synthetic division ( or long division) we obtain
x3 − 3x2 − 9x − 5 = (x − 5)(x2 + 2x + 1) = (x − 5)(x + 1)2 .
(c) Find all the zeros of P (x) and state the multiplicity for each zero.
Give the exact answer. No decimals.
x = 5 multiplicity 1.
x = −1 multiplicity 2.
(d) Find where P (x) approaches when x → ∞ and when
x → −∞.
P (x) → ∞ as x → ∞.
P (x) → −∞ as x → −∞.
(e) Sketch the graph of P (x) indicating the zeros.
20
15
10
5
x = −1
x=5
0
−5
−10
−15
−20
−25
−30
−35
−6
−4
−2
0
2
4
6
8
(10) Consider the polynomial P (x) = x(x + 2)2 (x − 3)3 .
(a) Find all zeros of P (x) and state their multiplicities.
x = −2 multiplicity 2.
x = 0 multiplicity 1.
x = 3 multiplicity 3.
(b) Where P (x) approaches when x → ∞ and when x → −∞.
P (x) → ∞ as x → ∞.
P (x) → ∞ as x → −∞.
(c) Sketch the graph of the function P (x) indicating the x-intercepts
(the zeros).
100
80
60
40
20
0
x=0
x=−2
x=3
−20
−40
−60
−80
−100
−4
−3
−2
−1
0
1
2
3
4
5
6
7
(11) Find the quotient (Q(x)) and the remainder (R(x)) using the long division.
3x4 − 5x3 + 4x + 3
x2 + x + 3
.
9
2
x
4
3
+x +3 3x −5x
3x4 +3x3
−8x3
−8x3
+3x2
+0x2
+9x2
−9x2
−8x2
−x2
−x2
−8x
+4x
−1
+3
+4x
−24x
+28x
−x
29x
+3
+3
−3
+6
Q(x) = 3x2 − 8x − 1.
R(x) = 29x + 6.
(12) An object is projected upward from the top of a building. The height of
the object in meters is described by the function h(t) = −4.9t2 + 40t +
10, where t is in seconds and corresponds to the moment the object is
projected.
(a) Determine the height of the building.
In the t = 0 moment the object is in the top of the building. So,
when t = 0 then h = 10 meters. Thus, the height of the building is
10 meters.
(b) Algebraically determine for what value of t the object reaches the
maximum height and determine this maximum height.(Hint use the
vertex formula.) Check your answer with your calculator.
Note that the graph of the height describes by the moving object
is an open downward parabola. So, the maximum height is at the
vertex of the parabola. Using the vertex formula with b = 40 and
b
a = −4.9 the object reaches the maximum height when t = − 2a
=
4.081632653 seconds. Substituting t = 4.081632653 seconds into
h(t) = −4.9t2 + 40t + 10. we obtain that the maximum height is
91.63265306 meters.
(c) Algebraically determine when the object reaches the ground. (Hint
use the quadratic formula). Check your answer with your calculator.
The object reaches the ground when h = 0. So, we have to substitute h = 0 and solve the equation for t using the quadratic formula
with c = 10, b = 40 and a = −4.9. Since the variable t describes
the time we just consider the positive solution. The solution for the
equation 0 = −4.9t2 + 40t + 10 is t = 8.406044918 seconds.
(13) The function p(x) = −x2 + 46x−360 models the daily profit in hundreds
of dollars for a company that manufactures x computers daily. (You may
show your work algebraically or graphically, which includes a sketch of
the graph.)
10
(a) How many computers should be manufactured each day to maximize profit?
Note that the graph of the function which describes the daily profit
is an open downward parabola. So, the maximum profit is at the
vertex of the parabola. Using the vertex formula with b = 46 and
a = −1 the manufacturer reaches the maximum profit when x =
b
− 2a
= 46 . So, x = 23 computers should be manufactured each day
to maximize profit.
(b) What is the maximum daily profit?
Substituting x = 46 into p(x) = −x2 + 46x − 360. we obtain that
the maximum profit is $ 16,900, since the profit function describes
the profit in hundreds of dollars.
(14) (a) Consider the rational function R(x) =
x2 − x − 6
.
2x2 − 2
(i) Find the vertical asymptotes of R(x) if there is any.
Factor the numerator and the denominator if you can.
(x − 3)(x + 2)
x2 − x − 6
=
2
2x − 2
2(x − 1)(x + 1)
If there is no common factor the vertical asymptotes of a
rational function are there where the denominator is 0.
So, the vertical asymptotes are x = 1 and x = −1.
(ii) Find the horizontal asymptote of R(x) if there is any.
If the degree of the numerator=the degree of the denominator then the horizontal asymptote is the ratio of the leading
1
coefficients. So, the horizontal asymptote is y = .
2
(iii) Find the x-intercepts of R(x) if there is any.
The x intercepts of a rational function are there where the
numerator is 0. So, the x intercepts are x = 3 and x = −2.
(iv) Find the y-intercept of R(x) if there is any. Substitute x = 0
(if you can) and solve it for y.
−6
= 3.
So, the y-intercept is y = −2
(b) Consider the rational function R(x) =
x+3
.
x2 + x − 2
(i) Find the vertical asymptotes of R(x) if there is any. Factor
the numerator and the denominator if you can.
x+3
x+3
=
2
x +x−2
(x − 1)(x + 2)
11
If there is no common factor the vertical asymptotes of a
rational function are there where the denominator is 0.
So, the vertical asymptotes are x = 1 and x = −2.
(ii) Find the horizontal asymptote of R(x) if there is any.
If the degree of the numerator<the degree of the denominator
then the horizontal asymptote is always 0. So, the horizontal
asymptote is y = 0.
(iii) Find the x-intercepts of R(x) if there is any.
The x intercepts of a rational function are there where the
numerator is 0. So, the x intercept is x = −3.
(iv) Find the y-intercept of R(x) if there is any.
Substitute x = 0 (if you can) and solve it for y.
3
So, the y-intercept is y = − .
2
x3 − x
(c) Consider the rational function R(x) = 2
.
x +1
(i) Find the vertical asymptotes of R(x) if there is any. Factor
the numerator and the denominator.
x3 − x
x(x − 1)(x + 1)
=
2
x +1
x2 + 1
If there is no common factor the vertical asymptotes of a
rational function are there where the denominator is 0. Since
there is no real solution for x2 + 1 = 0, there is no vertical
asymptote.
(ii) Find the horizontal asymptote of R(x) if there is any.
If the degree of the numerator>the degree of the denominator
then there is no horizontal asymptote.
(iii) Find the x-intercepts of R(x) if there is any.
The x intercepts of a rational function are there where the
numerator is 0. So, the x intercepts are x = 0, x = 1 and
x = −1.
(iv) Find the y-intercept of R(x) if there is any.
Substitute x = 0 (if you can) and solve it for y.
So, the y-intercept is y = 0.
(15) The number of deer in a state forest can be modeled using the the
rational function
6t2 + 1
)
3t2 + 1
where t is the time in years after the herd was first introduced.
N(t) = 4500(
12
(a) Approximately how many deer are there 5 years after they were
introduced into the forest?
6t2 + 1
Substitute t = 5 into the formula N(t) = 4500 2
. Approxi3t + 1
mately 8940 deer.
(b) In the long run, how many deer will there be in this state forest?
Explain the reasoning behind your answer. Also explain what possible real-world factors might limit the number of deer in the forest.
You have to find the horizontal asymptote of the rational function
6t2 + 1
N(t) = 4500 2
. The rational function approaches to the hori3t + 1
zontal asymptote as t increases. As t → ∞ N(t) → 4500 · 63 = 9000
deer. The answer is 9000 deer.
13
(16) Solve for x.
(a) ln x + ln(x − 15) = ln 34.
ln(x(x − 15)) = ln 34
x(x − 15) = 34
x2 − 15x − 34 = 0
(x + 2)(x − 17) = 0
Possible solutions are x = −2 and x = 17. Checking the possible
answers, we find that the only solution is x = 17.
(b) log x + log(x + 3) = 1.
log(x(x + 3)) = 1
The equivalent exponential form is
(x(x + 3)) = 10
x2 + 3x − 10 = 0
(x + 5)(x − 2) = 0
So, the solution is x = 2.
(c) e2x − 3ex − 10 = 0.
Substitute y = ex . Then the equation becomes
y 2 − 3y − 10 = 0
(y − 5)(y + 2) = 0
So, y = e = 5 or y = ex = −2. The solution for y = ex = 5 is
x = ln 5. There is no solution of y = ex = −2.
So, the solution is x = ln 5.
x
14
(17) (a) Use logarithms to find the solution correct to 3 decimal places.
22x−3 = 32x+1
Take the natural logarithm (ln) of both sides of the equation.
ln 22x−3 = ln 32x+1
Using the 3rd law of logarithm we obtain
(2x − 3) ln 2 = (2x + 1) ln 3
ln 2 = 0.693 and ln 3 = 1.099 with 3 decimal places accuracy. So,
(2x − 3)0.693 = (2x + 1)1.099
After distribution we have
1.386x − 2.079 = 2.198x + 1.099
Now solve the equation for x
−3.178 = 0.812x
x = −3.914
(b) Use the Laws of Logarithms to rewrite the following expression in
a form with no logarithm of a product, quotient or power.
r
y3
4
log (x
)
z2
r
3/2
y3
4y
log (x4
)
=
log
(x
) = log (x4 )+log (y 3/2 )−log z = 4 log x+3/2 log y−log z
z2
z
So, the answer is 4 log x + 3/2 log y − log z.
(c) Rewrite the following expression as a single logarithm.
2 log (x + 1) − 3 log (y − 5) + log z
2 log (x + 1)−3 log (y − 5)+log z = log (x + 1)2 −log (y − 5)3 +log z = log[
The answer is log[
(x + 1)2 · z
].
(y − 5)3
(x + 1)2 · z
]
(y − 5)3
15
(18) Find the time required for an investment of 2500 dollars to grow to 9000
dollars at an interest rate of 6.5 percent per year, compounded quarterly.
A(t) = P (1 + r/n)nt
P =$ 2500
A(t)=$ 9000
r = 0.065
n=4
Find t (number of years)
9000 = 2500(1 + 0.065/4)4t
dividing both sides of the equation by 2500
3.6 = (1 + 0.065/4)4t
3.6 = (1.01625)4t
taking the natural logarithm (ln) of both sides of the equation
ln 3.6 = ln((1.01625)4t)
using the 3rd Law of the Logarithm
ln 3.6 = (4t) ln(1.01625)
t=
ln 3.6
= 19.86636111 years
4 ln(1.01625)
(19) You are in a group of city planners that is trying to determine whether
or not to expand your water supply facilities. To aid in your decision,
you will use the information from the last two census figures for the city,
showing a population of 80,000 at the start of 1990 and a population
of 88,300 at the start of 2000. Knowing that you can currently supply
enough water for 95,000 people use the exponential model Q(t) = Q0 ·ekt
to determine during what year you will no longer have enough water to
meet the needs of your city.
When t = 0 then Q0 = 80, 000
When t = 10 then Q(10) = 88, 330
88, 330 = 80, 000ek10
1.104125 = ek10
ln 1.104125 = k10
ln 1.104125
= 0.009905316608 = k
10
16
The exponential model is
Q(t) = 80, 000 · e0.009905316608t
Find t when Q(t) = 95, 000
95, 000 = 80, 000 · e0.009905316608t
1.1875 = e0.009905316608t
ln 1.1875 = 0.009905316608t
ln 1.1875
= 17.349 = t
0.009905316608
So, 1990+17=2007, when the city no longer has enough water.
(20) Find the half-life of a radioactive substance if 200 grams of the substance
decays to 180 grams in 2 year.
Q(t) = Q0 · e−kt
Q0 = 200, when t = 2 Q(2) = 180. Find k.
180 = 200 · e−k2
0.9 = e−k2
ln 0.9 = −k2
ln 0.9
= 0.05268025783 = k
−2
Thus, the equation describing the procedure is
Q(t) = 200 · e−0.05268025783t
To find the half-life, substitute Q(t) = 100 and calculate t (the time
during the radio active material loses half of its original mass).
100 = 200 · e−0.05268025783t
1/2 = e−0.05268025783t
ln 1/2 = −0.05268025783t
ln 1/2
= 13.16 = t is the half-life.
−0.05268025783
17
(21) (a) Find the exponential function f (x) = ax whose graph goes through
the point. (2, 10.3).
Substitute x = 2 and y = 10.3 and solve the equation for a.
10.3 = a2
Note, the the equation above is not exponential. It is just a simple
quadratic equation. Taking the square root of both side of the
equation we get
p
a = (10.3) = 3.209361307
So, the solution is
y = 3.209361307x.
(b) For the logarithmic function log3 (x − 4) = y find the domain and
the x-intercept.
To find the domain of log3 (x − 4) = y, note that we can take the
logarithm only positive numbers. So, x − 4 > 0 that is x > 4. So
the domain is (4, ∞).
To find the x intercept substitute y = 0 and solve the equation
log3 (x − 4) = 0. The equivalent exponential form is x−4 = 30 = 1,
that is x=5. So, the x intercept is x = 5.
(22) Find all the solutions of the following system of equations.
(a)
8x − 3y = −3
5x − 2y = −1
We use the elimination method.
8x − 3y = −3 multiply by -2
Then, we get
5x − 2y = −1 multiply by 3
−16x + 6y = 6
15x − 6y = −3
Add the equations above. Then you eliminate the variable y.
−x = 3
x = −3
18
(b)
Substitute x = −3 either of the original equations and solve it for
y. We obtain y = −7.
So, the solution is x = −3 and y = −7.
x − 2y = 2
y 2 − x2 = 2x + 4
We use the substitution method. Let express x from the first equation.
x = 2 + 2y
Substitute into the second equation
y 2 − (2 + 2y)2 = 2(2 + 2y) + 4
Combining and rearranging the terms we obtain
3y 2 + 12y + 12 = 0
Simplifying by 3
y 2 + 4y + 4 = 0
which is
(c)
(y + 2)2 = 0
So, y = −2. From the first equation x = 2 + 2y, thus x = −2.
The solution is x = −2 and y = −2.
x + 4y = 8
3x + 12y = 2
We use the elimination method.
x + 4y = 8 multiply by -3
3x + 12y = 2
Then, we get
−3x − 12y = −24
3x + 12y = 2
Add the equations above. We obtain,
0 = −22
19
which is impossible. It means there is no solution of this system of
equations. Geometrically it means that the two lines represented
by the equations are parallel, so they never intersect.
(d)
3x2 + 4y = 17
2x2 + 5y = 2
We use elimination method.
3x2 + 4y = 17 multiply by -2
2x2 + 5y = 2 multiply by 3
We obtain,
−6x2 − 8y = −34
6x2 + 15y = 6
Add the equations above. Then we have,
7y = −28
y = −4
The substitute y = −4 into the first equation. We have
x2 = 11
√
√
So x = 11 and x = −√ 11.
√
The solutions are x = 11 and y = −4 or x = − 11 and y = −4