2.1 Graphs of Quadratic Functionsy = x2

x y0 01 1-1 12 4-2 4

A Parabola

Vertex

Lowest (Minimum) point ifThe parabola opens upward,And highest (Maximum) point if parabola opens downward.

Do you know what an axis of symmetry is?

Graphs of Quadratic Functions

y = x2 - 2

x y0 -21 -1-1 -12 2-2 2

Vertex

Notice this graph is shifted down 2 from the origin.Y = x2 – k (shifts the graph down k units)Y = x2 + k (shifts the graph up k units)

To shift the graph to the right or to the left y = (x – h)2 (shifts the graph to the right)

y = (x + h)2 (shifts the graph to the left)

Graphing Quadratic Equations y = ax2 + bx + c

a, b, c are real numbers & a 0

A quadratic Equation: y = x2 + 4x + 3 a = _____ b = _____ c = ______1 4 3

1) Where is the vertex? 4) What is the range?2) Where is the axis of symmetry? 5) What are the y-intercepts? (Does one always exist?)3) What is the domain? 6) What are the x-intercepts? (Does one always exist?)

Formula forVertex:

X = -b 2a

Plug x in toFind y

x y-2 -1 (Vertex)

-1 0-3 0 0 3-4 3 1 8-5 8

2

2

2

3)(

2)(

)(

xxf

xxf

xxf

2

2

2

3)(

2)(

)(

xxf

xxf

xxf

0a0a

The Role of “a” in y = ax2 + bx + c

Quadratic Equation Forms• General Form of a Quadratic:

• Standard (or Vertex) Form of a Quadratic:

khxay 2)(

cbxaxy 2

Vertex = (h, k)

Vertex = (-b/(2a) , f(-b/(2a)))

ExamplesFind the vertex, axis of symmetry, and graph each using symmetric picks or transformations

a.

b.

c.

2)5(3 2 xy

3)2(2 2 xy

6)4(2

1 2 xy

Vertex (5, 2)

Vertex (-2, -3)

Vertex (4, -6)

d. y = 2x2 –7x - 4

Vertex (7/4, -81/8) =

(1.75, -10.125)

Convert from Standard/Vertex Form to the General Form

Standard/Vertex Form: y = 2(x + 2)2 + 1

To change to standard form, perform multiplication, add, and combine like terms.

y = 2 (x + 2) (x + 2) + 1

y = 2 (x2 + 2x + 2x + 4) + 1

y = 2 (x2 + 4x + 4) + 1

y = 2x2 + 8x + 8 + 1

y = 2x2 + 8x + 9 (General Form)

Convert from General Form to Standard/Vertex Form

(Completing the Square – Example 1)

Example 1: y = x2 –6x – 1 (General Form) (b/2)2 = (-6/2)2 = (-3)2 = 9

y = (x2 –6x + 9) – 1 -9

y = (x – 3) (x – 3) – 1 – 9

y = (x – 3)2 – 10 (Vertex Form)

Step 1: Check the coefficient of the x2 term. If 1 goto step 2 If not 1, factor out the coefficient from x2 and x terms.

Step 2: Calculate the value of : (b/2)2

Step 3: Group the x2 and x term together, then add (b/2)2 and subtract (b/2)2

Step 4: Factor & Simplify

Convert from Standard Form to Vertex Form(Completing the Square – Example 2)

Example 2: y = 2x2 +4x – 1 (Standard Form) y = 2( x2 + 2x) –1 (2/2)2 = (1)2 = 1

y = 2(x2 +2x + 1) – 1 -2 (WHY did we subtract 2 instead of 1?)

y = 2(x + 1) (x + 1) – 1 – 2

y = 2(x + 1)2 – 3 (Standard/Vertex Form)

Step 1: Check the coefficient of the x2 term. If 1 goto step 2 If not 1, factor out the coefficient from the x2 and x term.

Step 2: Calculate the value of : (b/2)2

Step 3: Group the x2 and x term together, then add (b/2)2 and subtract (b/2)2

Step 4: Factor & Simplify

Solving Quadratic Equations General Form of a Quadratic Equation

y = ax2 + bx + c 0 = ax2 + bx + c (If y = 0, we can solve for the x-intercepts)

A quadratic Equation: y = x2 + 4x + 3 a = _____ b = _____ c = ______1 4 3

x y-5 8-4 3-3 0-2 -1-1 0 0 3 1 8

Formula forVertex:

X = -b 2a

Vertex (-2, -1)

NumericalSolution

Graphical SolutionSymbolic/Algebraic Solution

x2 + 4x + 3 = 0

(x + 3) (x + 1) = 0

x + 3 = 0 x + 1 = 0

x = -3 x = -1

Number of Solutionsy = x2 + 4x + 3

2 Real Solutions x = -3 x = -1

y = x2 - 4x + 4

1 Real Solution x = 2

y = 2x2 + 1

NO Real Solutions(No x-intercepts)

Quadratic Graphing Things to Know

General Form of a quadratic: y = ax2 + bx + c Standard/Vertex Form of a quadratic: y = a(x – h)2 + k

Using either form above:• Find and graph the vertex

• Use an x/y chart to plot points & graph the parabola

• Use and describe transformations to graph the parabola

• Use intercepts & symmetry while graphing the parabola.

• Show and give the equation for the line of symmetry.

• Convert from general to standard/vertex form and vice vera

Quadratic Modeling Application

Insurance Claims: The years 1999 to 2005 were particularly costly for insurance companies, with 7 of the 10 most costly catastrophes in U.S. history (as of 2005). For the United States Automobile Association (USAA) and its affiliates, the total cost of claims for catastrophic losses, in millions, can be approximated by C(x) = 34.87x2 – 98.1x + 258.3, where x = 0 for 1999, x = 1 for 2000, x = 2 for 2001, and so on.(a) Estimate the total cost of claims for the year 2003(b) According to the model during which year were catastrophic loss

claims at a minimum?(c) Would C(x) be useful for predicting total catastrophic loss claim for

the year 2015? Why or why not?

2.2 Polynomial Functions

Degree of a Polynomial: Largest Power of X that appears. The zero polynomial function f(x) = 0 is not assigned a degree.

** Polynomials of degree 3, 4, and 5 are named cubic, quartic, and quintic respectively

Polynomials are SMOOTH and CONTINUOUS.

Constant Term

Leading Term & Coefficient

Identifying Polynomial Functions

Power FunctionsA power function of degree n is a function of the form

a is a nonzero real number and n is a positive integer.

If n is even, then f(x) is an “even” function since axn = a(-x)n

• The graph is symmetric with respect to the y-axis• The graph of y = xn (where n is even) is similar to the graph of y = x2

For a > 0 : As x and As x For a < 0 : As x and As x

If n is odd, then f(x) is an “odd” function since a(-x)n = -axn

• The graph is symmetric with respect to the origin• The graph of y = xn (where n is odd) is similar to the graph of y = x3

For a > 0 : As x and As x For a < 0 : As x and As x

y = x2

x

y = x3

x

Leading Coefficient Test for End Behavior

• Even-degree polynomial functions have graphs with the same behavior at each end.

• Odd-degree polynomial functions have graphs with opposite end behavior.

• Note that the degree is even.

End Behavior : The behavior of a graph for large values of | x |

As x As x

As x As x

As x As x

As x As x

End Behavior Continued…

Zeros of Polynomial Functions

Example: Find a polynomial of degree 3 whose zeros are -4, -2, and 3 4 2 3f x x x x

4 2 3f x x x x

2 4 2 3f x x x x

A polynomial function of degree n with real coefficients, has at most, n real zerosFind all real zeros of the polynomial : f(x) = x3 + 2x2 – x – 2 and g(x) = x3 -2x2 + x - 2

Let a and b be two numbers with a < b. If f is a polynomial function such that f(a) and f(b) have opposite signs, then there is at least one number c, with a < c < b, for which f( c ) = 0. (The intermediate value theorem approximates a root by determining a very small range in which a root occurs.)

Example: Show that f(x) = x3 - 2x – 5 has a real zero between 2 and 3 Find an approximation to the nearest 10th.

A real zero exists if there is a sign change between f(2) and f(3)

f(2) = -1 (negative) Since one is negative and the other positive, there f(3) = 16 (positive) exists a real zero (x-intercept) between x = 2 and x = 3Find an approximation for the x-intercept between 2 and 3 to the nearest 10th.

Create a chart looking for the exact point of sign change.

x f(x) = x3 - 2x – 5 2 -1 2.1 .061

Sign change occurs between 2 and 2.1f(2.1) is closer to zero than f(2) so the x-interceptis approximated to be at x=2.1. For an estimate tothe nearest 100th repeat process between 2.0 and 2.1

Intermediate Value Theorem

3 42 2 1 3f x x x x

For the polynomial, list all zeros and their multiplicities.

Identifying Zeros and Their Multiplicities

2 With multiplicity 1-1 with multiplicyty 33 with multiplicity 4

Graphing a Polynomial Using it’s x-intercepts

(Local/Relative Maximum & Local/Relative Minimum)

A)

B)

C)

D)

For the polynomial

list the zero and its multiplicity and whether it touches or crosses the x axis at each x-intercept.

A)

B)

C)

D)

Which of the following polynomial functions might have the graph shown?

2.3 Review of Polynomial Long DivisionExample: Divide 4 – 5x – x2 + 6x3 by 3x – 2.

Begin by writing the divisor and dividend in descending powers of x. Then, figure out how many times 3x divides into 6x3.

3x – 2 6x3 – x2 – 5x + 42x2 Divide: 6x3/3x = 2x2.

6x3 – 4x2

Multiply.

Multiply: 2x2(3x – 2) = 6x3 – 4x2.

Divide: 3x2/3x = x.

Now, divide 3x2 by 3x to obtain x, multiply then subtract.

3x – 2 6x3 – x2 – 5x + 46x3 – 4x2

2x2 + x

3x2 – 5x

Multiply.

Multiply: x(3x – 2) = 3x2 – 2x.

3x2 – 2xSubtract 3x2 – 2x from 3x2 – 5x and bring down 4.

+ 4-3x

– 5x Subtract 6x3 – 4x2 from 6x3 – x2 and bring down –5x.

3x2

Subtract -3x + 2 from -3x + 4, leaving a remainder of 2.

2

-3x +2

-1

Answer: 2x2 + x – 1 + 2 3x - 2

Synthetic Division

To divide a polynomial by x – cExample

1. Arrange polynomials in descending powers, with a 0 coefficient for any missing terms. x – 3 x3 + 4x2 – 5x + 5

2. Write c for the divisor, x – c. To the right, 3 1 4 -5 5 write the coefficients of the dividend.

3. Write the leading coefficient of the dividend 3 1 4 -5 5 on the bottom row. Bring down 1.

1

4. Multiply c (in this case, 3) times the value 3 1 4 -5 5 just written on the bottom row. Write the 3 product in the next column in the 2nd row. 1

Multiply by 3.

5. Add the values in this new column, writing the sum in the bottom row.

6. Repeat this series of multiplications and additions until all columns are filled in.

7. Use the numbers in the last row to write the quotient and remainder in fractional form. The degree of the first term of the quotient is one less than the degree of the first term of the dividend. The final value in the row is the remainder.

3

1

4

-5

5

3

1

7

Add.

3

1

4

-5

5

3

21

48

1

7

16

53

Add.

Multiply by 3.

3

1

4

-5

5

3

21

1

7

16

Add.

Multiply by 3.

1x2 + 7x + 16 + 53x – 3

Synthetic Division

Division by x-k(Synthetic Division)

2 3 -8 7 -6 6 -4 6

3 -2 3 0

3x3 –8x2 + 7x – 6

x - 2

Answer: 3x2 –2x +3 remainder 0

Try this one:

x4 –5x3 + 9x2 –10x + 3

x - 3

3 1 -5 9 -10 3 3 -6 9 -3

1 -2 3 -1 0

Answer: x3 –2x2 +3x -1 remainder 0

The Remainder TheoremIf a polynomial f(x) is divided by x – a then the remainder, R = f(a)

x – 3 x3 + 4x2 – 5x + 5

Recall that with synthetic division, we found the remainder to be 53.

Now try f(a) = f(3) = (3)3 + 4(3)2 -5(3) + 5 = 27 + 4(9) -15 +5 = 27 + 36 -15 + 5 = 53

Example1:

3x3 –8x2 + 7x – 6

x - 2

Example2: Recall the remainder with synthetic division was 0 Now, try finding it with the “remainder theorem”

f(2) = 3(2)3 -8(2)2 +7(2) -6 = 3(8) -8(4) +14 -6 = 24 -32 +14 -6 = 0

The Factor Theorem: A polynomial f(x) has (x – a) as factor if and only if f(a) = 0.

Solving Polynomial Functions(Finding Zeros /x-intercepts w/Rational Zeros Test)

The rational zero test (P. 141) gives a method for finding the zeros (ie – x-intercepts) of a given polynomial equation.

Possible rational zeros = Factors of the constant term Factors of the leading coefficient

Example: f(x) = x3 + 8x2 + 11x - 20 [List all possible rational zeros]

Possible rational zeros = 1, -1, 2, -2, 4, -4, 5, -5, 10, -10, 20, -20 1, -1

Possible rational zeros = 1/1 -1/1 2/1 -2/1 4/1 -4/1, 5/1 -5/1, 10/1, -10/1 20/1, -20/1 = 1, -1, 2, -2, 4, -4, 5, -5, 10, -10, 20, -20

So, maybe: (x – 1) (x + 1) (x – 2) (x + 2) (x – 4) (x + 4) (x – 5) (x + 5) (x + 10) (x – 10) (x + 20) (x – 20) is a factor. We must test these with synthetic division!

Example Continued...Example: f(x) = x3 + 8x2 + 11x - 20 [Find the Zeros]

Possible rational zeros = 1, -1, 2, -2, 4, -4, 5, -5, 10, -10, 20, -20

Use synthetic division to test possible factors: (x – 1) (x + 1) (x – 2) (x + 2) (x – 4) (x + 4) (x – 5) (x + 5) (x –10)(x + 10) (x –20) (x + 20)

1 1 8 11 -20 1 9 20 1 9 20 0

After you find ONE factorwrite the factored polynomial & set = 0

Then, repeat the process until youare able to easily factor/solve.

(x – 1) (x2 + 9x +20) = 0(x – 1) (x + 4 ) (x +5 ) = 0

x – 1 = 0 x + 4 = 0 x + 5 = 0

x = 1 x = -4 x = -5

2.4 Descarte’s Rule of Signsf(x) is a polynomial function in descending order with real coefficients1. The number of positive zeros is the number of variations of sign of

f(x) or less than that number by an even integer.

2. The number of negative zeros is either equal to the number of variations of sign of f(–x) or less than that number by an even integer

** For Descarte’s Rule, a zero of multiplicity m is counted as m zeros.

Example: How many positive and negative zeros exist?

f x x5 3x3 5x2 9x 7. 3 Sign Changes, so, 3 positive zeros or 3-2 = 1

f x x 5 3 x 3 5 x 2 9 x 7

x5 3x3 5x2 9x 7 2 Sign Changes, so, 2 negative zeros or 2-2 = 0

2.4 Narrowing Down the Zeros(Making the pool of zeros smaller)

The list of possible zeros may be large. The upper and lower bound theorem/rules for bounds (P. 146) willhelp us limit the number of zeros we need to examine.

If a polynomial function f(x) has no zeros greater than a number k, then k is called an upper boundno zeros less than a number k, then k is called a lower bound

Let f(x) be a polynomial function w/real coefficients & positive leading coefficient.F(x) is synthetically divided by x – k.1. If k > 0 and each number in the last row is zero or positive, k is an upper bound2. If k < 0 and numbers in the last row alternate sign, then k is a lower bound

** Zeros in the last row can be regarded as positive or negative

Example: (Upper and lower bound theorem) Show that all the real roots of the equation 8x3 + 10x2 - 39x + 9 = 0 lie between –3 and 2.

2 is an upper bound if synthetic division of the polynomial by (x - 2) yields all non-negative numbers in the bottom row.

3513268

265216

9-391082All numbers in this row are nonnegative.

-3 is a lower bound if synthetic division of the polynomial by (x – (-3) ) yields ‘alternating signed’ numbers in the bottom row. (The number zero may beconsidered positive or negative)

03-148

-942-24

9-39108-3 Counting zero as negative, the signs alternate: +, -, +, -.

2.4 Complex ZerosRecall that a number z = a + bi is a complex number where a and b are real and

Fundamental Theorem of Algebra & Factorization Theorem for Polynomials (P. 148)In the complex number system, every nth-degree polynomial equation has exactly n roots and can be factored into exactly n linear factors.

Factorization Theorem for Polynomials with Real Coefficients (P. 150)Every polynomial with real coefficients can be uniquely factored over the real numbers as a product of linear factors and/or irreducible quadratic factors.

Conjugate Pairs Theorem (P. 149)If a polynomial function with real coefficients has a complex zero (a + bi),Then its conjugate (a – bi) is also a zero.

Odd Degree Polynomials with Real Zeros (P. 150)Any polynomial of odd degree with real coefficients must have at least one real zero.

ExamplesExample1Find a polynomial, P(x) of degree 4 with a leading coefficient of 2 and zeros –1, 3, i and –i.

Using the Factorization Theorem for Polynomials, since P(x) has degree 4,

1 2 3 4

1

2 1 3

32

P x x x x x

x x x x

x x x i x i

r r r

i i

a r

Example 2A polynomial P(x) of degree 9 with real coefficients has the following zeros: 2, of multiplicity 3; 4 + 5i, of multiplicity 2; and 3 – 7i. Write all nine zeros of P(x).

Using the Conjugate Pairs Theorem,We know 4-5i and 3+7i must be zeros.The 9 Zeros are:2, 2, 2, 4+5i, 4+5i, 4-5i, 4-5i, 3+7i, 3-7i

38© 2010 Pearson Education, Inc. All rights reserved

Example 3:Find all zeros of the polynomial P(x) = x4 – x3 + 7x2 – 9x – 18.

Solution

Possible zeros are: ±1, ±2, ±3, ±6, ±9, ±18

Use synthetic division to find that 2 is a zero.

(x – 2) is a factor of P(x). Solve

3 2

2

2

2

2

9 9 0

9 0

9 0

1 0 or 9 0

1 or 9

1 or

1

3

1

1

x x x

x

x

x

x x

x

x

x x

x x i

The four zeros of P(x) are –1, 2, –3i, and 3i.

39© 2010 Pearson Education, Inc. All rights reserved

Example 4:Given that 2 – i is a zero ofFind the remaining zeros.

Using the conjugate pairs theorem 2 – i and 2 + i are both zeros

So P(x) has linear factors:

2 2

2

2 1 4 5

1 1 4 5

1 1 2 2

P x x x x x

x x x x

x x x i x i

Therefore,

The zeros of P(x) are 1 (of multiplicity 2), 2 – i, and 2 + i.

2.5 Graphing Rational Functions

Rational functions are quotients of polynomial functions.The following are examples of rational functions. Can you find the domains? (All real numbers except values that make the denominator zero)

f(x) = x2 + 7x + 9 f(x) = x2 – 9 g(x) = x h(x) = x + 3

x(x – 2)(x + 5) (x – 3) (x2 – 9) x2 + 9

Arrow NotationSymbol Meaningx a + x approaches a from the right.x a - x approaches a from the left.x x approaches infinity (x increases without bound)x - x approaches negative infinity (x decreases without bound)

The line x = a is a vertical asymptote of the graph of a function f if f (x) increases or decreases without bound as x approaches a. (|f(x)| ∞ as x a+ or as x a-)f (x) as x a

+ f (x) as x a – f (x) as x a

+ f (x) - as x a -

f

a

y

x

x = a

fa

y

x

x = a

f (x) or f(x) - as x approaches a from either the left or the right.Vertical asymptotes occur at all x’s where the ‘simplified’ denominator is zero.

f

a

y

x

x = a

x = a

fa

y

x

The line y = b is a horizontal asymptote of the graph of a function f if f (x) approaches b as x increases or decreases without bound.

f

y

x

y = b

x

y

f

y = b f

y

x

y = b

f (x) b as x f (x) b as x f (x) b as x

Finding Horizontal Asymptotes

The degree of the numerator is n. The degree of the denominator is m.1. If n<m, the x-axis, or y=0, is the horizontal asymptote of the graph of f.

2. If n=m, the line y = an/bm is the horizontal asymptote of the graph of f.3. If n>m,the graph of f has no horizontal asymptote.Note : In contrast to vertical asymptotes, at most 1 horizontal asymptote exists and the graph may cross the horizontal asymptote.

Let:

Slant/Oblique Asymptotes

-2 -1 4 5 6 7 8321

7

6

5

4

3

1

2

-1

-3

-2Vertical asymptote: x = 3

Slant asymptote: y = x - 1

A slant asymptote exists if the deegree of the numerator is onemore than the degree of the denominator.

To find the equation of the slant asymptote, divide x - 3 into x2 - 4x – 5, and drop the remainder. The asymptote is given by the resulting linear equation.

The equation of the slant asymptote is y = x - 1.

Degree 2Degree 1

3 1 -4 -5 3 -3

1 -1 -8

Strategy for Graphing a Rational Functionf(x) = p(x) ,where p(x) and q(x) are polynomials with no common factor. q(x)

1. Determine whether the graph of f has symmetry. f (-x) = f (x): y-axis symmetry

f (-x) = -f (x): origin symmetry

2. Find the y-intercept (if there is one) by evaluating f (0).3. Find the x-intercepts (if there are any) by solving the equation p(x) = 0.

4. Find any vertical asymptote(s) by solving the equation q (x) = 0.5. Find the horizontal asymptote (if there is one)

6. Plot points between and beyond each x-intercept and vertical asymptote.7. Use the information obtained previously to graph the function.

Which of the following is the graph of the equation:

A) B)

C) D)

A)

B)

C)

D)

A)

B)

C)

D)

A)

B)

C)

D)