2.1 – Quadratic Functions and Models
Standard Form a.k.a. vertex form: f ( x)  a( x  h)2  k
a0
Vertical Parabola with an axis of x=h
Vertex: (h,k)
Opens up if a>0 and down if a<0
f ( x)  2 x 2  8 x  7
Graphing a parabola in standard form:
f ( x)  2( x 2  4 x  __)  7  __
f ( x)  2( x 2  4 x  4)  7  8
f ( x)  2( x  2) 2  1
Vertex is (-2, -1)
Finding the vertex and x-intercepts:
b
(this is the x-coordinate of the vertex). Substitute this
2a
value into the original equation to get the y-coordinate.
Either use vertex form or use: x  
f ( x)   x 2  6 x  8
f ( x)  ( x 2  6 x  8)
f ( x)  ( x  4)( x  2)
0  ( x  4)( x  2)
x  4 or 2
Write the equation of a parabola with a vertex of (1, 2) that passes through (0,0).
f ( x )  a ( x  h) 2  2
f ( x)  a( x  1) 2  2
0  a(0  h) 2  2
a  2
f ( x)  2( x  1) 2  2
2.2 – Polynomial Functions
Polynomial Functions are always continuous with smooth rounded turns. There are no breaks in
the graph of a polynomial function.
End Behavior:
If the highest exponent is even: the graph rises to the left and right
If the highest exponent is odd: the graph falls to the left and rises to the right
Notation:
f ( x)  
as x  
f ( x)   x3  4 x
Find the end behavior of the following functions: g ( x)  x 4  5 x 2  4
h( x )  x 5  x
If the leading coefficient is negative this reflects the end behavior about the x-axis.
Real Zeros of Polynomial Functions


The most real zeros a function can have is its degree (n).
The graph of the function can have at most n-1 turning points.
Equivalent Statements about Real Zeros
1.
2.
3.
4.
x=a is a zero of the function f
x=a is a solution of the polynomial equation f(x)=0
(x-a) is a factor of the polynomial f(x)
(a,0) is an x-intercept of the graph of f
Graph the following equation:
f ( x)  2 x 4  2 x 2
0  2 x 4  2 x 2
0  2 x 2  x 2  1
0  2 x 2 ( x  1)( x  1)
x  0,1,  1
Double Zeros (any even) will bounce off the axis
Single Zeros (any odd) will cross through the axis
Graph the following: f ( x)  ( x  1)( x  2)( x 2 )( x  3)( x  1)
f ( x)  2 x 3  6 x 2 
9
x
2

1
Graph the following: f ( x)   x 4 x 2  12 x  9
2
1
f ( x)   x(2 x  3) 2
2

Intermediate Value Theorem
f ( x)  x 3  x 2  1
x
f(x)
-2 -3
-1 1
0
1
1
3
There must be a zero between x=-2 and x=-1
In fact the function has one zero at about (-1.4656, 0)
2.3 – Polynomial Long Division
1
3
3 232
77
First divide using regular long division.
21
22
21
1
x 1 x2  4x  5  x  5
Then divide polynomials: x  1 x3  0 x 2  0 x  1  x 2  x  1
x 2  2 x  3 2 x 4  4 x3  5 x 2  3x  2  2 x 2  1 
2.3 – Synthetic Division & 2.4 Complex Numbers
To divide ax3  bx2  cx  d by x  k
k
a
b
c
d
ka
k(b+ka)
a b+ka
Continue the pattern to complete the synthetic division
Use synthetic division to divide: x 4  10 x 2  2 x  4 by x  3
-3
1
1
0
-10
-2
4
-3
9
3
-3
-3
-1
1
1
1x 3  3 x 2  1x  1 
1
x3
The Remainder and Factor Theorems


If a polynomial f(x) is divided by x-k, the remainder is r=f(k).
A polynomial f(x) has a factor (x-k) if and only if f(k)=0.
o (it goes in evenly if and only if the remainder is zero)
x 1
x  2x  3
2
Use the remainder theorem to evaluate the function at x=-2 if f ( x)  3x3  8 x 2  5 x  7
-2
3
3
8
5
-7
-6
-4
-2
2
1
-9
Because the remainder is -9, f(-2)=-9.
Show that (x-2) and (x+3) are factors of f ( x)  2 x 4  7 x3  4 x 2  27 x  18
2
-3
2
7
-4
-27
-17
4
22
36
18
2
11
18
9
0
2
11
18
9
-6
-15
-9
5
3
0
2
f ( x)  ( x  2)( x  3)(2 x 2  5 x  3)
f ( x)  ( x  2)( x  3)(2 x  3)( x  1)
i  1
i 2  1
i 3  i    1
i 26  i 2  1
i4  1
i88440  i 4  1
i 5  1
i 6  1
i 7  i    1
i8  1
Adding and Subtracting Complex Numbers
(4  7i)  (1  6i)  (4  1)  (7i  6i)
 5i
3i  (2  3i)  (2  5i)  (2  2)  (3i  3i  5i)
 0  5i
 5i
Multiplying Complex Numbers
(4  5i ) 2  (4  5i)(4  5i)
 16  20i  20i  25i 2
 16  40i  25
 9  40i
(4  5i )(4  5i )  16  25i
 16  25
 41
2
The second group are complex conjugates: (a+bi)(a-bi)
Write the following in standard form (a+bi)
2  3i  2  3i   4  2i 



4  2i  4  2i   4  2i 
8  4i  12i  6i 2

16  4i 2
8  6  16i

16  4
2  16i

20
1 4
  i
10 5
48  27  48i  27i
 4 3i  3 3i
3 

 3i

5 7  10  21  3 10i  7 5i  5 2

 
2.5 Rational Zeros
The Rational Zero Test:
f ( x)  an x n  an 1 x n 1  ...  a2 x 2  a1 x  a0
If the polynomial function has integer coefficients, then every rational zero has the form:
Rational Zero 
p
q
p = a factor of the constant term a0
q = a factor of the leading coefficient an
Example 1:
Find the rational zeros of:
f ( x)  x 4  x 3  x 2  3x  6
p 1,  2,  3,  6

q
1
Possible Rational Zeros :  1,  2,  3,  6

 21  5 2  7 5  3 10 i
By applying synthetic division, you can determine that x=-1 and x=2 are the only two rational
zeros.
-1
2
1
-1
1
-3
-6
-1
2
-3
6
1
-2
3
-6
0
1
-2
3
-6
2
0
6
0
3
0
1

f ( x)  ( x  1)( x  2) x 2  3

There are only two real zeros: x=-1 and x=2
Example: 2
List all of the possible rational zeros and then factor the function:
g ( x)  2 x 3  3 x 2  8 x  3
Possible Rational Zeros:
1
2
2

Factors of 3 1,  3
1 3

 1,  3,  , 
Factors of 2 1,  2
2 2
3
-8
3
2
5
-3
5
-3
0
g ( x)  ( x  1) 2 x 2  5 x  3

g ( x)  ( x  1)(2 x  1)( x  3)
Example 3:
Find all of the real solutions of the equation: 10 x3  15 x 2  16 x  12  0
The constant term is -12
The leading coefficient is -10
Possible Rational Solutions:
1,  2,  3,  4,  6,  12
1,  2,  5,  10
With so many possibilities, use your graphing calculator to sketch a graph and determine some
good guesses
.
It looks like the graph passes through x=2
Using synthetic division, we will check that solution.
2
-10
-10
15
16
-12
-20
-10
12
-5
6
0
( x  2)(10 x 2  5 x  6)  0
Using the quadratic formula for the second factor:
5  (5) 2  (4)(10)(6)
x
2( 10)
5  25  ( 240)
20
5  265
x
20
x
The three real solutions are: 2, -1.0639, 0.5639
Conjugate Pairs
All complex zeros occur in conjugate pairs. If a+bi is a zero of the function, then a-bi is also a
zero of the function.
Example 4:
Find a fourth degree polynomial with real coefficients that has zeros of: -1, -1, 3i
f ( x)  ( x  1)( x  1)( x  3i )( x  3i )

f ( x)   x

 2 x  1 x
f ( x)  x 2  2 x  1 x 2  9i 2
2
2
9


f ( x)  x 4  2 x3  x 2  9 x 2  18 x  9
f ( x)  x 4  2 x3  10 x 2  18 x  9
Example 5:
Find all of the zeros of f ( x)  x 4  3x 2  6 x 2  2 x  60 given that 1+3i is a zero of f.
All complex zeros come in conjugate pairs, thus, we also know that 1-3i is a zero.
 x  (1  3i) x  (1  3i)
( x  1)  3i ( x  1)  3i 
( x  1) 2  9i 2
x2  2x  1  9
x 2  2 x  10
Using long division, we can divide:
x2  x  6
x  2 x  10 x  3x  6 x  2 x  60
2

f ( x)   x
4
3
2


 2 x  10  ( x  3)( x  2)
f ( x)  x 2  2 x  10 x 2  x  6
2
x  1  3i,3,  2
Descartes’ Rule of Signs
The number of positive real zeros is equal to the number of sign changes in f(x) or less by an
even integer.
The number of negative real zeros is equal to the number of sign changes in f(-x) or less by an
even integer.
Example 6:
Find the number of each type of zero:
f ( x)  3x3  5 x 2  6 x  4
f ( x)  3( x)3  5( x) 2  6( x)  4
f ( x)  3 x 3  5 x 2  6 x  4
This function has three variations in sign for f(x) and zero variations in sign for f(-x)
This function could have either three positive real zeros or one positive real zero and no negative
real zeros. From the graph, you can see that the function only has one positive real zero near
x=1. This means that it will have two complex zeros.
Upper and Lower Bound Rules
If f(x) is divided by x-c using synthetic division:


If c>0 and each number in the last row is either positive or zero, c is an upper bound for
the real zeros of f.
If c<0 and the numbers in the last row are alternately positive and negative (zero entries
count as either), c is a lower bound for the real zeros of f.
Look at example 1:
-1 was the lower bound and 2 was the upper bound
Example 7:
f ( x)  2 x3  3x 2  23x  12
Find the real zeros of p 1,  2,  3,  4,  6,  12
1
3

  ,  1,  ,  2,  3,  4,  6,  12
q
1,  2
2
2
.5
2
2
-3
-23
+12
1
-1
-12
-2
-24
0
1

f ( x)   x   2 x 2  2 x  24 2
2

1

f ( x)  2  x   x 2  x  12
2




1

f ( x)  2  x   ( x  4)( x  3)
2

Zeros: 0.5, 4, 3

2.7 – Nonlinear Inequalities
Polynomial Inequalities
1. Critical Numbers: Find all the real zeros of polynomials and arrange them in increasing
order.
2. Use the critical numbers to determine test intervals.
3. Choose one x-value in each interval to evaluate the polynomial at. The value of the
polynomial at the test point is representative of the entire interval.
Example 1:
Solve the Polynomial Inequality
x2  5x  6  0
( x  1)( x  6)  0
The critical numbers are x=-1 and x=6.
The test intervals are: (, 1), (1, 6), and (6, )
Interval
( ,  1)
(1, 6)
(6, )
Test Value
-2
0
10
Polynomial Value
8
-6
44
The solution to the inequality is: (-1, 6)
Example 2:
Solve the Polynomial Inequality
3x3  4 x 2  12 x  16
3x3  4 x 2  12 x  16  0
( x  2)( x  2)(3x  4)  0
The critical numbers are x=2, x=-2 and x=
4
3
Conclusion
Positive
Negative
Positive
4 4 

The test intervals are: (, 2),  2,  ,  , 2  , and (2, )
3 3 

Interval
(,  2)
4

 2, 
3

4 
 ,2
3 
(2, )
Test Value
-3
-1
Polynomial Value
-65
21
3
2
3
7
8
25

Conclusion
Negative
Positive
Negative
Positive
4

The solution to the inequality is:  2,  and (2, )
3

If the original inequality had been: 3x3  4 x 2  12 x  16 we would have used brackets instead
4

of parenthesis:  2,  and [2, )
3

Example 3:
Unusual Solution Sets:
A. x 2  6 x  9  0
The solution set is all real numbers: (, )
This is because the quadratic is positive for all values of x.
B. x 2  4 x  4  0
The solution set is only a single real number: {2}
This is because the vertex of the parabola is at (2, 0) and this is the only point less than or
equal to zero.
C. x 2  5 x  7  0
The solution set is empty. The quadratic is not less than zero for any value of x.
D. x 2  6 x  9  0
The solution set consists of all real numbers except x=3. Interval notation: (,3) and
(3, ) .
Solving a Rational Inequality
Rational Expressions can change signs at its zeros and at its undefined values (vertical
asymptotes). These occur at the zeros of the numerator and the zeros of the donominator.
Example 4
Solve:
2x  7
3
x5
2x  7
3 0
x 5
2 x  7  3 x  15
0
x5
x  8
0
x 5
Critical Numbers: x=5 and x=8
The test intervals are: (,5), (5,8), and (8, )
Interval
Test Value
(,5)
0
(5,8)
(8, )
6
10
Polynomial Value
8

5
2
2

5
The solution to the inequality is: (,5) and [8,  )
Note the bracket at 8 because x can equal 8 but not 5.
Conclusion
Negative
Positive
Negative
Example 5
Solve:
3x  5
1
x 3
3x  5
1  0
x 3
3x  5  x  3
0
x 3
2x  2
0
x 3
Critical Numbers: x=1 and x=3
The test intervals are: (,1), (1,3), and (3, )
Interval
Test Value
( ,1)
0
(1,3)
(3, )
2
5
Polynomial Value
2
3
-2
4
Conclusion
Positive
Negative
Positive
The solution to the inequality is: [1,3)
Note the bracket at 1 because x can equal 1 but not 3.
Example 6
A projectile is fired straight up into the air with an initial velocity of 640 feet per second.
How long will it take to hit the ground?
When will it be more than 4000 feet off the ground?
s  16t 2  v0t  s0