Graphing Quadratic Functions in Standard Form Consider the function f(x) = 2x2 – 4x + 5. a. Determine whether the graph opens upward or downward. Because a is positive, the parabola opens upward. b. Find the axis of symmetry. The axis of symmetry is given by . Substitute –4 for b and 2 for a. The axis of symmetry is the line x = 1.

Graphing Quadratic Functions in Standard Form Consider the function f(x) = 2x2 – 4x + 5. c. Find the vertex. The vertex lies on the axis of symmetry, so the x-coordinate is 1. The y-coordinate is the value of the function at this x-value, or f(1). f(1) = 2(1)2 – 4(1) + 5 = 3 The vertex is (1, 3). d. Find the y-intercept. Because c = 5, the intercept is 5.

a = 2, so a > 0 Graph of f opens up. Vertex is a minimum. Finding the vertex and axis of symmetry by hand. a = 2, so a > 0 Graph of f opens up. Vertex is a minimum. f(x) = 2x2 + 2x – 1.5 Formula for finding the axis of symmetry: vertex: (–0.5, –2) axis of symmetry: x = –0.5

Write an equation for a parabola with vertex (–1, 1), opens up, and contains (2, 12 ). y = a (x – h)2 + k 12 = a (2 – (-1))2 + 1 12 = a(9) + 1 11 = 9a a = 11/9 y = 11/9 (x + 1)2 + 1

Leading coefficient is positive. Degree is odd and has 2 turns Identify and describe the important features of the graph of a polynomial function. g(x) = x3 – 6x2 + 9x Leading coefficient is positive. Degree is odd and has 2 turns End Behavior: Graph falls on left and rises on right.

Writing the function given the roots Write a polynomial function given sufficient information. zeros: 2, -4, 7 factors of P(x): (x – 2), (x + 4), and (x - 7) P(x) = (x – 2)(x + 4)(x - 7)

Determine all of the zeros = x 2(x + 3)(x + 2) Polynomial Function with Repeated Zeros Determine all of the zeros = x 2(x + 3)(x + 2) Answer: It has three zeros, x = 0 multiplicity of 2, x = –2, and x = –3.

Divide x 3 – x 2 – 14x + 4 by x 2 – 5x + 6. (–)x 3 – 5x 2 + 6x 4x 2 – 20x + 4 (–)4x 2 – 20x + 24 –20 Answer:

= Multiply by c, and write the product. Synthetic Division A. Find (2x 5 – 4x 4 – 3x 3 – 6x 2 – 5x – 8) ÷ (x – 3) using synthetic division. 3 2 –4 –3 –6 –5 –8 = add terms. 6 6 9 9 12 2 2 3 3 4 4 = Multiply by c, and write the product. coefficients of depressed quotient remainder

Answer:

Using Synthetic Division to Divide by a Linear Binomial Divide using synthetic division. (3x4 – x3 + 5x – 1) ÷ (x + 2) Step 1 Find a. a = –2 For (x + 2), a = –2. Step 2 Write the coefficients and a in the synthetic division format. 3 – 1 0 5 –1 –2 Use 0 for the coefficient of x2.

Example Continued Step 3 Bring down the first coefficient. Then multiply and add for each column. –2 3 –1 0 5 –1 Draw a box around the remainder, 45. –6 14 –28 46 3 –7 14 –23 45 Step 4 Write the quotient. 3x3 – 7x2 + 14x – 23 + 45 x + 2 Write the remainder over the divisor.

P is a polynomial function with integer coefficients. If is a root of P(x) = 0, then p q p is a factor of the constant term of P. q is a factor of the leading coefficient of P.

possible rational roots: Don’t repeat values 1  , 3 2 4 Find all possible rational roots of a polynomial equation. 3x3 – x2 + 12x – 4 = 0 P= -4 and q= 3 Factors of -4 are: 1, 2, 4 Factors of 3 are : 1, 3 possible rational roots: Don’t repeat values 1  , 3 2 4

Use synthetic substitution to verify that 2 + 5i is a zero of p (x). Find the Zeros of a Polynomial When One is Known Find all complex zeros of p (x) = x 4 – 6x 3 + 35x 2 – 50x – 58 given that x = 2 + 5i is a zero of p. Then write the linear factorization of p (x). Use synthetic substitution to verify that 2 + 5i is a zero of p (x). 2 + 5i 1 –6 35 –50 –58 (2 + 5i)(–4 + 5i) = –8 – 10i + 25i 2 2 + 5i –33 – 10i = –8 – 10i + 25(–1) 1 –4 + 5i = –33 – 10i

Find the Zeros of a Polynomial When One is Known 2 + 5i 1 –6 35 –50 –58 (2 + 5i)(2 – 10i) = 4 – 10i – 50i2 2 + 5i –33 – 10i 54 – 10i = 4 – 10i – 50(–1) 1 –4 + 5i 2 – 10i = 54 – 10i 2 + 5i 1 –6 35 –50 –58 (2 + 5i)(4 – 10i) = 8 – 50i 2 2 + 5i –33 – 10i 54 – 10i 58 = 8 – 50(–1) 1 –4 + 5i 2 – 10i 4 – 10i 0 = 58

Find the Zeros of a Polynomial When One is Known Because x = 2 + 5i is a zero of p, you know that x = 2 – 5i is also a zero of p. Divide the depressed polynomial by 2 – 5i. 2 – 5i 1 –4 + 5i 2 – 10i 4 – 10i 2 – 5i –4 + 10i –4 + 10i 1 –2 –2 0 Using these two zeros and the depressed polynomial from this last division, you can write p (x) = [x – (2 + 5i)][x – (2 – 5i)](x 2 – 2x – 2).

Identifying All of the Real Roots of a Polynomial Equation Identify all the real roots of 2x3 – 9x2 + 2 = 0. Step 1 Use the Rational Root Theorem to identify possible rational roots. ±1, ±2 = ±1, ±2, ± . 1 2 p = 2 and q = 2 Step 2 Graph y = 2x3 – 9x2 + 2 to find the x-intercepts. The x-intercepts are located at or near –0.45, 0.5, and 4.45. The x-intercepts –0.45 and 4.45 do not correspond to any of the possible rational roots.

Step 3 Test the possible rational root . Example 4 Continued 1 2 Step 3 Test the possible rational root . 1 2 Test . The remainder is 0, so (x – ) is a factor. 1 2 2 –9 0 2 1 –4 –2 2 –8 –4 The polynomial factors into (x – )(2x2 – 8x – 4). 1 2 Step 4 Solve 2x2 – 8x – 4 = 0 to find the remaining roots. 2(x2 – 4x – 2) = 0 Factor out the GCF, 2 Use the quadratic formula to identify the irrational roots. 4± 16+8 2 6 2 x = = ±

f (x) = a[x – (–1)](x – 2)[x – (2 – i)](x – (2 + i)] Find a Polynomial Function Given Its Zeros Write a polynomial function of least degree with real coefficients in standard form that has –1, 2, and 2 – i as zeros. Because 2 – i is a zero and the polynomial is to have real coefficients, you know that 2 + i must also be a zero. Using the Linear Factorization Theorem and the zeros –1, 2, 2 – i, and 2 + i, you can write f (x) as follows: f (x) = a[x – (–1)](x – 2)[x – (2 – i)](x – (2 + i)]

f (x) = (1)(x + 1)(x – 2)[x – (2 – i)][x – (2 + i)] Let a = 1. Find a Polynomial Function Given Its Zeros While a can be any nonzero real number, it is simplest to let a = 1. Then write the function in standard form. f (x) = (1)(x + 1)(x – 2)[x – (2 – i)][x – (2 + i)] Let a = 1. = (x2 – x – 2)(x2 – 4x + 5) Multiply. = x4 – 5x3 + 7x2 + 3x – 10 Multiply. Therefore, a function of least degree that has –1, 2, and 2 – i as zeros is f (x) = x4 – 5x3 + 7x2 + 3x – 10 or any nonzero multiple of f (x). Answer: Sample answer: f (x) = x4 – 5x3 + 7x2 + 3x – 10

Set the top=0 Set the bottom=0

Some rational functions have a horizontal asymptote Some rational functions have a horizontal asymptote. The existence and location of a horizontal asymptote depends on the degrees of the polynomials that make up the rational function.

Identify the zeros and asymptotes of the function. x – 2 x2 – 1 f(x) = x – 2 (x – 1)(x + 1) f(x) = Zero: 2 Vertical asymptote: x = 1, x = –1 Horizontal asymptote: y = 0

Find Vertical and Horizontal Asymptotes A. Find the domain of and the equations of the vertical or horizontal asymptotes, if any. Answer: D = {x | x ≠ 1, x }; vertical asymptote at x = 1; horizontal asymptote at y = 1

Key Concept 3 You must use long division to find the oblique asymptote

Identify the zeros, y-int and asymptotes of the function. Then graph. 3x2 + x x2 – 9 f(x) = x(3x – 1) (x – 3) (x + 3) f(x) = Zeros: 0 and – 1 3 Vertical asymptote: x = –3, x = 3 Horizontal asymptote: y = 3 Y-int: (0, 0)

The values of h and k affect the locations of the asymptotes, the domain, and the range of rational functions whose graphs are hyperbolas.

y-int (0, -5/3) and x –int (-5/2, 0) Identify the asymptotes, domain, and range of the function g(x) = – 2. 1 x + 3 Vertical asymptote: x = –3 Domain: {x|x ≠ –3} The value of h is –3. Horizontal asymptote: y = –2 y-int (0, -5/3) and x –int (-5/2, 0) The value of k is –2.