Download Math 120 Exam 5 Solutions to the Study Guide Use pencil. Simplify

Math 120
Exam 5
Solutions to the Study Guide
Use pencil. Simplify all solutions. Show your work. Clearly identify your final solutions.
No calculators.
1. Solve each quadratic equation by completing the square.
a)
section 8.1 (problem 55)
b)
8.2 (8)
3x 2  7 x  2  0
x 2  73 x   23
49
49
x 2  73 x  36
  32  36
 x  76 
2

25
36
x  76   56
x  2, 13
2. Solve each quadratic equation by using the quadratic formula.
a)
8.2 (19)
b)
8.2 (8)
3x 2  7 x  2  0
x

7  49  4  3 2 
2  3
75
6

7  25
6
 2, 13
3. Solve each quadratic equation using any of the methods established in class.
a)
8.1 (25)
(Square Root Rule)
b)
8.1 (35)
(Square Root Rule)
c)
solve
8.2 (41)
7
7

1
x x4
x  5  53
4. Solve the equations for all solutions.
a)
x 4  8 x 2  20  0
x
2
or x 2  2  0
x 2  10
or
or
x 3  x 3  20  0
2
x
 10  x 2  2   0
x 2  10  0
x   10
b)
1
1
3

x5 x 6  0
c)


5 x 3  4  0
x 3 5  0
1
1

x 3 40
1
or
x 2  2
3
x 5
or
x  i 2
x  125
or
3
x 6


x  6  0 or
x  4

x 1  0

x  6 or
x  64

x 1  0
x  1
x  36 only
5. Solve each inequality. Use interval notation to display the solutions.
a)
 2 x  5 x  4   0
b)
x2  4
c)
x3  5x 2  6 x  0
Step 1: Solve for x (as if it were an equation).
x  x  6  x  1  0
x  4, 52
x  1,0,6
x  2
Step 2: Test ‘x’ values and list solutions as intervals.
 4, 52 
 , 2   2,  
 1,0  6,  
6. Write a quadratic function in the form f ( x)  ax 2  bx  c that has the given roots. (Use proper notation.)
a)
x = –3 and
7
2
f ( x)   x  3 2 x  7 
 f ( x)  2 x 2  x  21
b)
x  2  5 (undo the complete the sq. process.)
x  2 5
 x2  5
  x  2  5
2
 x2  4x 1  0
 f  x   x2  4 x  1
7. Change each function into vertex form.
a)
y  x 2  6 x  10
x – coordinate of vertex
x
b
2a
=
6
2
b)
x – coordinate of vertex
3
x
y - coordinate of vertex
b
2a

4
 4
2  12 
y - coordinate of vertex
y   3  6  3  10  1
y  12  4   4  4   7  1
2
2
Vertex form of the function
Vertex form of the function
y   x  3  1
y
2
8. Graph each function using the method established in class.
“We have reviewed function graphs in class.”
9.
y  12 x 2  4 x  7
Give an example of each of the following.
a)
radical expression
x5
b)
quadratic function
f ( x)  x 2  x  12
c)
rational expression
3x  1
x2
d)
linear function
y  2x  4
e)
quadratic equation
x2  2 x  4  0
f)
rational equation
3x  1
5
x2
g)
radical function
f ( x)  3 x  4
1
2
 x  4
2
1
10.
Graph the function and solve the equation.
a)
x2  4 x  6  0
b)
Solve: (I will use complete the square.)
f ( x)  x 2  6 x  5
Graph shown below: 1st find vertex.
b
2a
x coordinate: x 
x 2  4 x  4  6  4
y coordinate: y  4 (substitute x into the function)
 x  2
vertex form:
2
 2
= x
 6 
21
x 2  4 x  6
=3
y   x  3  4
2
x  2   2
x  2  i 2
11.
Given the function f ( x )  5 x 2  4 , answer the following.
a)
b)
List the domain and range of f(x). (use interval notation)
Domain:
 ,  
Range:
 4,  
2
2


f (2  h)  f (2) 5(2  h )  4   5  2   4 


h
h
5(4  4h  h 2 )  4   16
=
h
(You must show the correct substitution 1st.)
 20  20h  5h 2  4   16 20h  5h 2 h  20  5h 


 20  5h

h
h
h
12. Suppose that a flare is launched upward with an initial velocity of 80 ft/sec from a height of 224 ft. Its
h(t )  16t 2  80t  224 .
height in feet, h(t) after t seconds is given by:
a)
Find the height of the object after 2 seconds.
h(2)  320 ft.
b)
How many seconds will it take the object to hit the ground?
h(t )  0 (height = 0. Solve for t.)
16t 2  80t  224  0
divide by -16
t 2  5t  14  0
factors
 t  7  t  2   0
t  7 (only the positive solution)
c)
When will it reach its maximum height? How high will the flare go?
This parabola opens downward. The coordinates of the vertex (t, h(t)) will give you the time and the height.
b
2a
t
=
80 5
80


2  16  32 2
2.5 seconds until it reaches the vertex (highest point).
Height = h  52   16  52   80  52   224  16  254   200  224  100  200  224  324 ft.
2
The flare will reach its highest point in 2.5 seconds. The maximum height will be 324 ft.
13.
Section 8.4
problems similar to 1 – 10
Example:
A car travels 200 miles averaging a certain speed. If the car had traveled 10 miles per
hour faster, the journey would have taken 1 hour less. Find the average speed of the original trip.
rate problem
Rate
Rate * Time = Distance
*
Time
=
Distance
‘x’ mph
‘t’ hours
200 miles
‘x + 10’ mph
(t – 1) hours
200 miles (same dist.)
1st equation:
x  t  200
2nd equation:
 x  10   t  1  200
which can be written as t 
200
x
which can be written as t  1 
200
x  10
or t 
200
1
x  10
Since we have two equivalent expressions for time, we can therefore write the equation:
200
200

1
x
x  10
(Multiply to ‘clear’ the fractions.) …by x  x  10 
200  x  10   200 x  x  x  10 
(Now solve for x.)
200 x  2000  200 x  x 2  10 x
0  x 2  10 x  2000
zero product property
0   x  50  x  40 
x  40 mph (speed cannot be negative)
Check:
40 mph for 5 hours takes us 200 miles
50 mph for 4 hours takes us 200 miles (10mph faster in 1 hour less) 
14.
Section 8.8
problems similar to 7 – 9, 13
Example:
The revenue (in dollars) made by selling ‘x’ boxes of an item can be determined by
the function R( x )  9.5 x  0.05 x 2 . How many boxes must be sold to maximize the revenue?
1st:
R(x) is a quadratic function that can be written as:
R( x )  0.05 x 2  9.5 x
Note: This is a quadratic function (parabola) that opens down.
It will have a maximum value at the vertex.
2nd:
Solution
To find the vertex, x 
x
9.5
9.5

2  0.05 0.1
b
2a
:
 95 (Multiply by ‘1’ which is
10
10
in this case.)
The maximum revenue is earned when 95 boxes of the item are sold.
15.
List the 4 methods for solving a quadratic equation.
Done in Class!!!!!