Math 1200
Exam 3A (Fall 2013) Solutions
Name: _________
Show all your work in order to get full credit. Each question (or sub–question) is worth 5
points.
1. Assume that the cost C is linearly related to output x. A company that manufactures
roller skates has fixed costs (cost at 0 output) of $300 per day and total costs of $4300
per day at an output of 100 pairs of skates per day. Find the equation of the line relating
output to cost.
The two points are (0, 300) and (100, 4300).
Slope =
= 40
Equation of the required line through two points (0, 300) and (100, 4300)
with slope 40 is
y – 300 = 40(x – 0)
y= 40x + 300
2. Write the equation of the line passing through (–8, –4) and perpendicular to the line
1
6
defined by y  x  3 .
Slope of the line perpendicular to the given line is –6.
So, equation of the line passing through (–8, –4) with slope –6 is
y − (– 4) = – 6(x – (– 8))
y + 4 = – 6(x + 8)
y = – 6x – 52
3. Let g ( x)  3 x  2  4 .
a. What is the parent function? Parent function is x
b. Describe (in words) the order of transformations from the parent function to g(x).
Move the parent function x 4 units down,2 units to the right and stretch
vertically by a factor of 3.
c. Use the following grid to graph g ( x)  3 x  2  4.
4. Determine if the function m( x)  5 x 5  x 3 is even, odd, or neither. Show your work.
m(x) is odd if m( x)  m( x)
m( x)  5( x) 5  ( x) 3
5
3
Since m( x)  5 x  x =  m(x)
So, m(x) is an odd function
5. The graph of y  f (x) is given below.
a. Graph y   f (x)
b. Y= f(-x)
6. Let f(x) be defined by the following multi–part function.
 x  7 for x  2

f ( x)   x 2 for  2  x  1
 3 for 1  x  4

a. Find. f (3)
b. Find f (2) .
For x < –2, f(x) = x + 7
For  2  x  1 , f(x) = x 2
So, f(–3) = – 3 + 7 = 4
7. Find the difference quotient
So, f(–2) = (2) 2 = 4
f ( x  h)  f ( x)
if f ( x)  y  x 2  4 .
h
Since f ( x)  x 2  4
So,
f ( x  h)  f ( x) ( x  h) 2  4  ( x 2  4)

h
h
x 2  2 xh  h 2  4  x 2  4 2 xh  h 2
h( 2 x  h )



 2x + h
h
h
h
8. Let f ( x)  3x  4 and g ( x) 
a.
1
.
x 1
Find  f o g (x) = f(g(x))= f (
1
3
3  4x  4 4 x  1
+4=
)

x 1
x 1
x 1
x 1
b. Find g o f ( x) = g(f(x)) = g(3x + 4) 
1
1

(3x  4)  1 3( x  1)
c. Indicate the domain of ( f  g )(x) and ( g  f )(x) , respectively.
Domain of ( f  g )(x) is all real numbers, but x ≠ 1
Domain of ( g  f )(x) is all real numbers, but x≠-1.
9. Given h( x)  3 x  5 , find two functions, f and g , such that h( x)  ( f  g )(x) .
The inner function, g(x) = x + 5 and the outer function, f(x) =
3
x
10. Write f ( x)  3x 2  6 x  1 in vertex form.
f ( x)  3(( x 2  2 x  1)  1))  1
f ( x)  3( x  1) 2  3  1
f ( x)  3( x  1) 2  2
a. Identify the vertex. (1, –2)
b. Identify the axis of symmetry. x = 1
c. Find the minimum/maximum value and where it is achieved.
Since a = 3 > 0, parabola opens upward and the minimum value is – 2 and is achieved
at the vertex (1, –2).
11. A coin, thrown upward at time t = 0 from an office in the Empire State Building, has
height in feet above the ground t seconds later given by h(t )  16t 2  64t  960 .
a. From what height is the coin thrown?
At t= 0, h(0)  16(0) 2  64(0)  960 h(0)  960
b. At what time does the coin reach the ground?
Coin reaches the ground when h(t) = 0
0  16t 2  64t  960
 16(t 2  4t  60)  0
(t – 10)(t + 6) = 0
So, the only solution is t = 10
The coin reaches ground after ten minutes.