1 of 18 Pre-Cal Chapter 2 Section 1 SAT/ACT Warm - up

2 of 18 Pre-Cal Chapter 2 Section 1 Chapter 2 – Vocabulary Words 1.Polynomial function 2.Linear function 3.Quadratic function 4.Continuous 5.Intermediate Value Theorem 6.Synthetic division 7.Imaginary number 8.Complex conjugates 9.Rational function 10.Slant Asymptote See Page 172

Graphing Quadratic Functions Chapter2 Sec 1 Chapter 2 Sec 1

4 of 18 Pre-Cal Chapter 2 Section 1 Essential Question How do you sketch graphs and write equations of parabolas? Key Vocabulary: Polynomial Function Linear equation Quadratic function

5 of 18 Pre-Cal Chapter 2 Section 1Definitions

6 of 18 Pre-Cal Chapter 2 Section 1 Quadratic function The graph of a quadratic function is a parabola. Every parabola is symmetrical about a line called the axis (of symmetry). The intersection point of the parabola and the axis is called the vertex of the parabola. x y axis f (x) = ax 2 + bx + c vertex

7 of 18 Pre-Cal Chapter 2 Section 1

8 of 18 Pre-Cal Chapter 2 Section 1 Leading Coefficient The leading coefficient of ax 2 + bx + c is a. When the leading coefficient is positive, the parabola opens upward and the vertex is a minimum. When the leading coefficient is negative, the parabola opens downward and the vertex is a maximum. x y f(x) = ax 2 + bx + c a > 0 opens upward vertex minimum x y f(x) = ax 2 + bx + c a < 0 opens downward vertex maximum

9 of 18 Pre-Cal Chapter 2 Section 1 5 y x Simple Quadratic Functions The simplest quadratic functions are of the form f (x) = ax 2 (a  0) These are most easily graphed by comparing them with the graph of y = x 2. Example: Compare the graphs of

10 of 18 Pre-Cal Chapter 2 Section 1 Example: f(x) = (x –3) Example: Graph f (x) = (x – 3) and find the vertex and axis. f (x) = (x – 3) is the same shape as the graph of g (x) = (x – 3) 2 shifted upwards two units. g (x) = (x – 3) 2 is the same shape as y = x 2 shifted to the right three units. f (x) = (x – 3) g (x) = (x – 3) 2 y = x x y 4 4 vertex (3, 2)

11 of 18 Pre-Cal Chapter 2 Section 1 Completing the Square 1.Group together the x’s 2.Factor out the a. 3.Take coefficient of x. a.Divide by 2 b.Then square result 4.Add the “squared result” inside the parenthesis and subtract the “Squared result times a” outside the parenthesis. 5. Factor the parenthesis, it will be (x ± (3a.)) 2.

12 of 18 Pre-Cal Chapter 2 Section 1 x y Quadratic Function in Standard Form Example: Graph the parabola f (x) = 2x 2 + 8x + 5 and find the axis and vertex. f (x) = 2x 2 + 8x + 5 original equation f (x) = 2( x 2 + 4x) + 5 group the x’s & factor out 2 f (x) = 2( x 2 + 4x + 4) + 5 – 8 complete the square f (x) = 2( x + 2) 2 – 3 standard form a > 0  parabola opens upward like y = 2x 2. h = –2, k = –3  axis x = –2, vertex (–2, –3). x = –2 f (x) = 2x 2 + 8x + 5 (–2, –3)

13 of 18 Pre-Cal Chapter 2 Section 1 x y 4 4 Vertex and x-Intercepts Example: Graph and find the vertex and x-intercepts of f (x) = –x 2 + 6x + 7. f (x) = – x 2 + 6x + 7 original equation f (x) = – ( x 2 – 6x) + 7 factor out –1 f (x) = – ( x 2 – 6x + 9) complete the square f (x) = – ( x – 3) standard form a < 0  parabola opens downward. h = 3, k = 16  axis x = 3, vertex (3, 16). Find the x-intercepts by solving –x 2 + 6x + 7 = 0. (–x + 7 )( x + 1) = 0 factor x = 7, x = –1 x-intercepts (7, 0), (–1, 0) x = 3 f(x) = –x 2 + 6x + 7 (7, 0)(–1, 0) (3, 16)

14 of 18 Pre-Cal Chapter 2 Section 1 y x f (x) = a(x – h) 2 + k standard form f (x) = a(x – 2) 2 + (–1)vertex (2, –1) = (h, k) y = f(x) (0, 1) (2, –1) Since (0, 1) is a point on the parabola: f (0) = a(0 – 2) 2 – 1 1 = 4a –1 and Example: Find an equation for the parabola with vertex (2, –1) passing through the point (0, 1).

15 of 18 Pre-Cal Chapter 2 Section 1 Example: Find the vertex of the graph of f (x) = x 2 – 10x f (x) = x 2 – 10x + 22 original equation a = 1, b = –10, c = 22 At the vertex, So, the vertex is (5, -3). The vertex of the graph of f (x) = ax 2 + bx + c (a  0) Vertex of a Parabola

16 of 18 Pre-Cal Chapter 2 Section 1 Minimum and Maximum Values of Quadratic Functions

17 of 18 Pre-Cal Chapter 2 Section 1 The path is a parabola opening downward. The maximum height occurs at the vertex. At the vertex, So, the vertex is (9, 15). The maximum height of the ball is 15 feet. Example: A basketball is thrown from the free throw line from a height of six feet. What is the maximum height of the ball if the path of the ball is:

18 of 18 Pre-Cal Chapter 2 Section 1 Essential Question How do you sketch graphs and write equations of parabolas?

19 of 18 Pre-Cal Chapter 2 Section 1 Daily Assignment Chapter 2 Section 1/ Ch 1 Review Text Book (TB) Textbook Pgs 99– 102 #1 – 4, 7, 11, 17, 21, 27, 29, 31, 35, 55, 63 – 67 All R Read Sections 2.2 – 2.3 Show all work for credit.