Download Unit 3 Review for Common Assessment

Unit 3 Review for
Common
Assessment
Match the graph of a quadratic function with it’s
equation below:
f(x) = x2
f(x) = -(x+2)2+4
f(x) = (x+2)2-1
Describe the end behavior of the graph of each
given graph.
x , f (x) 
x , f (x) 


x , f (x) 
x , f (x) 
x , f (x) 
x , f (x) 


Use the Leading Coefficient Test to determine the end behavior of the graph of
the given polynomial function.
1.) f(x) = -x3 + 4x
Rise Left, fall right
x  , f (x)  
x  , f (x)  
EVEN
2.) f(x) = x4 – 5x2 +4
Rise left, rise right

x  , f (x)  
x  , f (x)  
3.) f(x) = x5 - x x , f (x) 
Fall left, rise right
4.) f(x) = x3 – x2 - 2x
Fall left, rise right
x , f (x) 
x , f (x) 
x , f (x) 
5.) f(x) = -2x4 + 2x2
Fall
left, fall right
x , f (x) 
x , f (x) 
Determine without graphing, the critical points
of each function.
1.) f(x) = (x + 2)2 - 3
f’(x) = -2x + 6
f’(x) = 2x + 4
Min (-2,-3)
3.) f(x) = 3x3 - 9x + 5
f’(x) = 9x2 - 9
2.) f(x) = -x2 + 6x - 8
f’’(x) = 18x
Max (3,1)
4.) = x3 + 6x2 + 5x
Min (-.47, -1.13)
Max (-3.53, 13.12)
Pt. of Inflection (-2,6)
Min ( 1, -1)
Max (-1, 11)
Pt. of Inflection ( 0 , 5)
Min ( -√5, -16)
4
2
5.) f(x) = x - 10x + 9
Max (0, 9)
Min ( √5 , -16)
Find the zeros of each polynomial function.
1.) x2 – 40 = 0
x 2  40
x(x 2  4 x  4)  0
x  40
x(x  2)(x  2)  0
x  2 10
2 + 11x – 102 = 0
3.)
x



2.) x3 + 4x2 + 4x = 0
(x 17)(x  6)  0
x = -17, 6
If you can’t figure it out then use

Quadratic Formula
x = 0, -2, -2
4.) x2 + ¾x + ⅛ = 0
x  12x  14 0
x = -½, -¼
Find the zeros of the polynomial
function by factoring.
1.) f(x) = x3 + 5x2 – 9x - 45
1.) f(x) = x3 + 4x2 – 25x - 100
x 2 (x  4)  25(x  4)  0
(x 2  25)(x  4)  0
(x  5)(x  5)(x  4)  0

x = 5, -5, -4
Which of the following is a rational zero of
f(x) = –2x5 + 6x4 + 10x3 – 6x2 – 9x + 4
1, -3, -2, 4, -1 ????
Remember you could use synthetic division or just do p(x) and see if you get
a remainder of ZERO
OR
p(4)  2(4) 5  6(4) 4 10(4) 3  6(4) 2  9(4)  4
So 4 is a factor, the others are not
=0
Use synthetic division to divide x4 + x3 – 11x2 – 5x + 30 by x - 2 . Then
divide by x + 3 Use the result to find all zeros of f(x).
So you are left with:
x2
-5
x2 x
C
Then all the zeroes are: -3,  5 , 2
R
List all possible rational
zeros of
1.)
2.)
List all possible rational roots, use synthetic division to find an actual
root, then use this root to solve the equation.
f(x) = 2x4 + x3 – 31x2 – 26x + 24
Hint 4 and -3/2 are roots
2x2 + 6x – 4
USE QUADRATIC FORMULA!!!
Find the number of possible positive,
negative, and imaginary zeros of:
2,0 positive roots
P
N
2
0
I
1 positive root
0
f (x)  x 2  4 x  5
f (x)  x 4  2x 3  x 2  2x  2
0
0
2
0 negative roots
P
3,1 negative roots
N
I
1
3
0
1
1
2

3,1 positive roots
f (x)  6x 4  x 3  4 x 2  x  2
1 positive root
P
N
I
3,1 positive roots
3
1
0 f (x)  5x 5  6x 4  24 x 3  20x 2  7x  2
1
1
2

2,0 positive roots
P
3
3
1
1
N
2
0
2
0
I
0
2
2
4
Use the given root to find the solution
set of the polynomial equation.
p(x) = x4 + x3 – 7x2 – x + 6
GIVEN -3 IS A ROOT
Then we can find the rest by factoring:
3
2
x  2x  x  2
x 2 (x  2) 1(x  2)
(x 2 1)(x  2)
(x 1)(x 1)( x  2)
So the roots are:
-3, -1, 1, and 2
Which equation represents
the graph of the function?
f(x) = 2x2+2x-1
f(x) = -x2-3x+4
f(x) = x2+10x-1
Approximate the real
zeros of each function.
R(x)  3x 4  x 2 1
0.7, -0.7
F(x)  x 3  4 x  6

H(x)  2x 3  4x 2  3
2.3
-2.5
G(x)  x 2  3x 1

-0.4 and -2.6
Use the given root to find the solution set
of the polynomial equations
x 4  x 3  8x 2  4 x  48
x 4  3x 3 12x 2  54 x  40
2i
Since 2i is a root, so is -2i
Turn the roots into factors, multiply

them together, then use long
division
(x  2i)(x  2i)  x  4
2
x  x 12
x 2  4 x 4  x 3  8x 2  4 x  48


2
Then factor to find the remaining roots
x  x 12  (x  3)(x  4)
2

So the roots are: 2i, -2i, 3, and -4
Since 3-i is a root, so is 3+i
3-i
Turn the roots into factors, multiply
them together, then use long
division
(x  (3  i))(x  (3  i))  x 2  6x 10
x 2  3x  4
x 2  6x 10 x 4  3x 3 12x 2  54 x  40

Then factor to find the remaining roots
x 2  3x  4  (x 1)(x  4)
So the roots are: 3-i, 3+I, 1, and -4
Find the vertical asymptotes, if any, of
the graph of each function.
R(x) 
x
x2  4
F(x) 

x = -2, x = 2
x3
x 4
x=4
x2  9
G(x)  2
x  4 x  21
x2
H(x)  2
x 1
No vertical asymptote

x = -7
Find the horizontal asymptote,
if any, of the graph of
R(x) 
x
x2  4
F(x) 

y=0
y=1
3x 4  x 2
G(x)  3
x  x 2 1
x2
H(x)  2
x 1
y=1
x3
x 4
If a monomial is on
bottom then you just
break it up.
Otherwise must do long division

3x  3
x 3  x 2 1 3x 4  0x 3  x 2  0x  0

y = 3x + 3
Choose the correct graph
for the rational function
x 2 1
2x 2  5x  2
x2
R(x) 
F(x) 
H(x)  2
x
x2  4
x 1



2(x  2) 2 (x  5)
G(x) 
(x  5)(x  2)2