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Math Test 2B Solutions
C. HECKMAN TEST 2B
SOLUTIONS 170
(1) [10 points] For the rational function below, find its
xintercept(s), yintercept(s), vertical
asymptotes, horizontal asymptotes, and determine whether it is even, odd, or
neither. (If the function does not have any vertical asymptotes (for example),
make
sure you answer “none”; do not leave your answer blank.)
Solution: Its x intercept is where y = 0, or when
, which is
when the numerator is zero . The numerator factors into 3(x + 4)(x − 1), so the
xintercepts are −4 and 1 .
The yintercepts are where x = 0. If you substitute x = 0 into the formula,
you have to divide by zero , so this function has no yintercept .
The vertical asymptotes occur where the denominator is zero but the numerator
isn’t. Since the denominator is zero when x^{3}(x + 5) = 0, the vertical
asymptotes are x = 0 and x = −5 .
To find the horizontal asymptote, we look at the degrees of the numerator
and denominator. Since the degree of the denominator is greater, there is a
horizontal asymptote of y = 0 .
Since , which is neither the original
function nor its
negative , this function is neither even or odd.
Grading: +2 points for each answer. Partial credit: +1 point for each answer.
Do the following for the equation 2^{y} = 5:
(a) [5 points] Convert it into logarithmic form .
Solution: .
Grading for partial credit: +2 points total for taking logarithms of both sides.
(b) [10 points] Solve for y, rounding your answer to three decimal places .
Solution: You can take logarithms of both sides here, and solve for y, or use
the
change of basis formula:
Note that will also
give you the correct answer.
(3) Solve the following equations for x . You must give exact answers!
(a) [10 points] ln(x + 1) + ln(2x − 1) = 3
Solution: Since there are multiple logarithms, you need to combine them :
Now convert to exponential form :
so you need to solve the quadratic equation
[You got 4 points for getting this far.] The solutions to
this quadratic equation
are [7 points], but only the solution with
the + sign makes
the original equation true. So the only solution is
. (This is not
3, since e^{3} is not exactly 20 but only approximately 20.)
Grading for partial credit: +3 points for miscellaneous work.
(b) [10 points]
Solution: If you let y = 10^{x}, you find out that
[3 points] so y = 1 or y = 4 [7 points]. To find x, you
solve the equations 10^{x} = 1
(which has a solution of 0) and 10^{x} = 4 (which has a solution of log 4). Both
solutions work in the original equation, so the solutions are 0 and log 4
.
Grading for partial credit: +7 points for 4 and 1. Points were taken off for
an approximate answer.
(4) A bank offers a savings account where the interest is compounded 5.1%
quarterly.
(a) [5 points] If you deposit $300 now, how much money will you have in 4 years?
Solution: Use the compound interest formula:
Grading: +2 points for the formula, +2 points for
substituting, +1 point for
the final answer. Grading for partial credit: −2 points for
.
(b) [10 points] When will you have $400 in your account?
Solution: You need to solve the equation
[3 points]
To solve this equation, you have to start by dividing both sides by 300; the
lefthand side is not 303.825^{4t}.
[7 points]
[10 points]
Grading for partial credit: −3 points for not dividing by 300 first.
(5) [10 points] Find a polynomial p (x) which has degree 4,
has 2 as a root, has −1 as a
root of multiplicity 3, and where p(1) = −24.
Solution: Since you know information about the roots (zeros), it is best to work
with the factored form. Since −1 has multiplicity 3, (x + 1)^{3} is a
factor of p(x),
and (x−2) is also a factor of p(x). Note that if the multiplicity of 2 were
greater
than 1, then p(x) would have degree at least 5. Hence p(x) = A(x − 2)(x + 1)^{3}
for some constant A.
If you use the condition p(1) = −24, you find out that
so A = 3. Thus .
Grading for partial credit: +3 points (total) for involving (x − 2)(x + 1);
+7 points (total) for (x − 2)(x + 1)^{3}.
(6) [10 points] Find the quotient and remainder when
3x^{4}−7x^{2}+8x+5 is divided by x^{2}+2.
Solution: Since the divisor is not of the form x − c, you must use long
division:
The quotient is 3x^{2} − 13 , and the remainder is 8x + 31 .
Grading was done on a 0–3–5–7–10 point basis.
(7) [10 points] Using the Rational Root Theorem, list the
possible rational roots of the
polynomial 3x^{4} −8x^{2} +5x−4. You do not need to determine which are actual roots.
Solution: The Rational Root Theorem states that all rational roots of this
polynomial
are of the form , where p is a factor of −4,
and q is a factor of 3. Since
the factors of −4 are 1, 2, 4, −1, −2, and −4, and the factors of 3 are 1, 3,
−1,
and −3, we take all possible combinations to get the list below.
Grading: +3 points for the factors of −4, +3 points for
the factors of 3, and
+4 points for combining them. Grading for common mistakes: +8 points (total)
if the reciprocals were given.
(8) [10 points] The rational roots of the polynomial
are −1 and. Find all
roots of p(x) exactly, along with their multiplicities.
Solution: First of all, you should divide p(x) by x − (−1), and then divide that
quotient by , to try to find an equation that
the other roots must satisfy.
Doing this division (by long division or synthetic division ) yields f(x) =
14x^{3} + 24x^{2} + 4x − 6.
You can’t use the quadratic formula on f(x), since f(x) has degree three.
The instruction to find the multiplicities suggests that −1 or
may have multiplicity
greater than one, so you should divide f(x) by x+1 and again, to see
whether one of these divisions produces a remainder of 0. In fact, x + 1 divides
into f(x) without a remainder. The quotient of this final division is
14x^{2}+10x−6,
and we can use the quadratic formula to find the remaining roots:
.
The roots of p(x) are thus −1 (with multiplicity 2, since you divided by x+1
twice), , and
.
Grading: +5 points for finding f(x), +2 points for dividing again by x + 1,
+3 points for using the quadratic formula. Grading for partial credit: +5 points
(total) for −1 and .
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