MAT 170 Test Solutions - Spring 2009, Exams of Pre-Calculus

Download MAT 170 Test Solutions - Spring 2009 and more Exams Pre-Calculus in PDF only on Docsity!

MAT 170

T.Miller

Test #2 FORM A

March 23 or 24, 2009

KEY

This test had 17 problems and 9 pages.

1. Find all solutions to the equation

x^3 + x^2 + 4x + 4 = 0.

Solution: First we note that all possible rational roots are ± 1 , ± 2 , ±4. All of the coefficients are positive hence there will not be any postive roots. So, we start by checking to see if − 1 is a root using synthetic division.

Now we only have to solve the quadratic x^2 + 4 = 0

x^2 + 4 = 0 =⇒ x^2 = − 4 =⇒ x = ± 2 i. So: x = − 1 , − 2 i, 2 i is the correct answer

2. Form a polynomial function with real coefficients that has degree 4, leading coefficient −3, with zeros at 4 of multiplicity 2, and − 3 i of multiplicity 1. Solution: The coefficients are real hence 3i is a root of multiplicity 1 because it is the conjugate of the root − 3 i of multiplicity 1. Every root, c, yields a factor x − c, so we have the factors (x − 4), (x + 3i), and (x − 3 i). The multiplicity of the root is the power of the factor and the leading coefficient is −3. Putting this all together, we get: P (x) = −3(x − 4)^2 (x + 3i)(x − 3 i). is the correct answer This has degree 4 so it is the polynomial we seek.
3. Use the properties of logarithms to expand the logarithmic expression as much as possible:

log

x^9 z^3 w^7 y^5

Solution: First apply the property loga

(m n

= loga m − loga n:

log(x^9 z^3 ) − log(w^7 y^5 ). Next apply the property loga(mn) = loga m + loga n:

log(x^9 ) + log(z^3 ) − (log(w^7 ) + log(y^5 )) = log(x^9 ) + log(z^3 ) − (log(w^7 ) − log(y^5 ). Now apply the property loga(mp) = p loga m:

9 log x + 3 log z − 7 log w − 5 log y is the correct answer

7. Find the domain of the function

h(x) = ln(12 − 3 x).

Solution: 12 − 3 x > 0 =⇒ 12 > 3 x =⇒ 4 > x So (−∞, 4) is the correct answer

prob Form a rational function with the following characteristics:

Vertical asymptotes at x = 6 and x = − 4 Horizontal asymptote at y = 2 x-intercepts at (− 1 , 0) and (2, 0)

Solution: The vertical asymptotes are zeros of the denominator, hence the denominator contains at least (x − 6)(x + 4). Since we have a non-zero horizontal asymptote, the denom- inator and numerator have the same degree and the ratio of the leading coefficients of the numerator and denominator is 2. Now we know that the function has the form

f (x) =

2(x − a)(x − b) (x − 6)(x + 4)

Finally the x-intercepts are zeros of the numerator, giving that the numerator has at least (x + 1)(x − 2). Thus we know the rational function:

f (x) =

2(x + 1)(x − 2) (x − 6)(x + 4)

is the correct answer

8. Write the equation in its equivalent exponential form:

log 5 (x − 3) = y.

Solution: Using the definition, 5y^ = x − 3 is the correct answer

9. Divide x^4 − 2 x^2 + 7 x − 2

Solution: Since the dividend is of the form x − c, we can use synthetic division.

So, x^3 + 2x^2 + 2x + 4 +

x − 2

is the correct answer

10. What is the multiplicity of the zero at x = −4 for the function

g(x) = − 8 x^4 (x − 2)^5 (x + 4)^3.

Solution: The zero −4 corresponds to the factor (x + 4) which is raised to the 3rd^ power. Hence, 3 is the correct answer

11. Which of the following is false?

Solution: 4 i(3 − 5 i) = 12i − 20 i^2 = 12i − 20(−1) = 20 + 12I 2 3 − i

3 − i

3 + i 3 + i

6 + 2i 9 + 1

2(3 + i) 10

3 + i 5 (7 − 3 i)(− 2 − 5 i) = − − 14 − 35 i + 6i + 15i^2 = 14 − 29 i + 15(−1) = 14 + 15 − 29 i = 29 − 29 i (4i)^2 = 16i^2 = 16(−1) = − 16

So, (4i)^2 = 16 is the correct answer

14. If the half-life of a radioactive element X is 5 seconds and there is initially 128 grams of X, after 16 seconds how many grams of X are present? Round answer to two decimal places.

Solution: First we will need to use the half-life to find the relative decay rate k. For y = y 0 ekt, we have y = 12 y 0 and t = 5. This gives

1 2

y 0 = y 0 e^5 k^ (dividing by y 0 ) 1 2

= e^5 k^ (rewriting as a logarithm)

ln

= 5k (dividing by 5)

k =

− ln 2 }

Now that we have k, we simply need to put y 0 = 128 and t = 16:

y = 128e16(−^ ln 2/5)^ =⇒ y = 13.93 grams

15. A person standing close to the edge on top of 200 foot building throws a baseball vertically upward. The quadratic function, s(t) = − 16 t^2 + 64t + 200, models the ball’s height, in feet, above the ground t seconds after it was thrown. After how many seconds does the ball reach its maximum height? Round to the nearest second, if necessary.

Solution: The ball will reach maximum height at the vertex of the parabola; to find the vertex, complete the square.

s(t) = − 16 t^2 + 64t + 200 s(t) = −16(t^2 − 4 t) + 200 s(t) = −16(t^2 − 4 t + 2^2 − 22 ) + 200 s(t) = −16(t^2 − 4 t + 2^2 ) − 22 (−16) + 200 s(t) = −16(t − 2)^2 + 64 + 200 s(t) = −16(t − 2)^2 + 264

The vertex is (2, 264), so the time to reach the maximum height is 2 seconds.

16. For the function f (x) = 2x^2 + 4x − 11 , use the technique of completing the square to put the quadratic in standard form and then find the vertex. (Hint: f (x) = a(x − h)^2 + k)

Solution:

s(t) = 2x^2 + 4x − 11 s(t) = 2(x^2 + 2x) − 11 s(t) = 2(x^2 + 2x + 2^2 − 22 ) − 11 s(t) = 2(x^2 + 2x + 2^2 ) − 22 (2) − 11 s(t) = 2(x + 2)^2 + 8 = 11 s(t) = −2(x + 2)^2 − 3

So the vertex is (− 2 , −3).