C. HECKMAN TEST 2A
SOLUTIONS 170
(1) [10 points] For the rational function below, find its
x-intercept(s), y-intercept(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 2(x − 2)(x − 3), so
the
x-intercepts are 2 and 3 .
The y-intercepts are where x = 0, so you just substitute x = 0 into the
formula to get or.
The vertical asymptotes occur where the denominator is zero but the numerator
isn’t. Since the denominator is never zero, there are no vertical asymptotes .
To find the horizontal asymptote, we look at the degrees of the numerator
and denominator. Since both degrees are equal, take the ratio of the
coefficients
of x2: or y = 2 .
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.
2) Do the following for the equation 4z = 19:
(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 z , rounding your answer to three decimal places.
Solution: You can take logarithms of both sides here, and solve for z, 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]
Solution: Since there are multiple logarithms, you need to combine them :
Now convert to exponential form :
[You got 4 points for getting this far.] If you
cross-multiply, you find out that
x + 37 = 10(x + 1) [7 points], so the only possible solution is 3 . It turns out
to
be an actual answer, after checking.
Grading for partial credit: +3 points for miscellaneous work.
(b) [10 points] 102x − 10x − 12 = 0
Solution: If you let y = 10x, you find out that
[3 points] so y = 4 or y = −3 [7 points]. To find x, you
solve the equations
10x = 4 (which has a solution of log 4) and 10x = −3 (which has no solutions).
So log 4 is the only solution.
Grading for partial credit: +7 points for 4 or log(−3). Points were taken off
for an approximate answer.
(4) A bank offers a savings account where the interest is compounded 4.7%
monthly.
(a) [5 points] If you deposit $350 now, how much money will you have in 5 years?
Solution: Use the compound interest formula:
Grading: +2 points for the formula, +2 points for
substituting, +1 point for
the final answer.
(b) [10 points] When will you have $800 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 350; the
left-hand side is not 351.3712t.
[7 points]
[10 points]
Grading for partial credit: −3 points for not dividing by 350 first.
(5) [10 points] Find a polynomial p (x) which has degree 3,
has 2 as a root, has −2 as a
root of multiplicity 2, and where p(0) = 16.
Solution: Since you know information about the roots (zeros), it is best to work
with the factored form. Since −2 has multiplicity 2, (x + 2)2 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 4. Hence p(x) = A(x − 2)(x + 2)2
for some constant A.
If you use the condition p(0) = 16, you find out that
so A = −2. Thus .
Grading for partial credit: +3 points (total) for involving (x − 2)(x + 2);
+7 points (total) for (x − 2)(x + 2)2.
(6) [10 points] Find the quotient and remainder when 4x4
+3x3 −4 is divided by x2 −2x.
Solution: Since the divisor is not of the form x − c, you must use long
division:
The quotient is , and
the remainder is .
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 9x4 +5x3 −7x+5. 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 5, and q is a factor of 9. Since
the factors of 5 are 1, 5, −1, and −5, and the factors of 9 are 1, 3, 9, −1, −3,
and −9, we take all possible combinations to get the list below.
Grading: +3 points for the factors of 5, +3 points for the
factors of 9, 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 −2 and . Find all
roots of p(x) exactly, along with their multiplicities.
Solution: First of all, you should divide p(x) by x − (−2), 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) = 14x3 + 38x2 + 34x + 28.
You can’t use the quadratic formula on f(x), since f(x) has degree three.
The instruction to find the multiplicities suggests that −2 or
may have multiplicity
greater than one, so you should divide f(x) by x+2 and
again, to see
whether one of these divisions produces a remainder of 0. In fact, x+2 divides
into
f(x) without a remainder. The quotient of this final division is 14x2 + 10x +
14,
and we can use the quadratic formula to find the remaining roots:
.
The roots of p(x) are thus −2 (with multiplicity 2, since you divided by x+2
twice),, and
.
Grading: +5 points for finding f(x), +2 points for dividing again by x + 2,
+3 points for using the quadratic formula. Grading for partial credit: +5 points
(total) for −2 and .