Warm-up: Sketch y = 3|x – 1| – 2 HW: page 214 (1-12, 27- 30, 37- 40, 47-50)

Graphing Quadratic Functions Objective: Graph a quadratic function Write a quadratic function in standard form Find the equation of a parabola

Definition of a Polynomial Function Let n be a nonnegative integer and let be real numbers with The function given by is called a polynomial function of x with degree n. f(x) = 9 degree of 0 constant function f(x) = 9x + 5 degree of 1 linear function f(x) = 9x2 + 5x + 1 degree of 2 quadratic function f(x) = 9x3 + 5x2 + x + 1 degree of 3 cubic function

The graph of a quadratic function is a parabola. Every parabola is symmetrical about a line called the axis of symmetry. x y f (x) = ax2 + bx + c vertex Axis of symmetry Vertex the intersection point of the parabola and the axis of symmetry

The leading coefficient of ax2 + bx + c is a. y a > 0 opens upward When the leading coefficient is positive (+) f(x) = ax2 + bx + c vertex minimum x y When the leading coefficient is negative (-) vertex maximum f(x) = ax2 + bx + c a < 0 opens downward

The simplest quadratic functions are of the form f (x) = ax2 (a  0) a < 1 wider a > 1 narrower These are most easily graphed by comparing them with the parent function y = x2. Example: Compare the graphs of , and 5 y x -5 g(x) = 2x2

Example: Graph f (x) = (x – 3)2 + 2 and find the vertex and axis. f (x) = (x – 3)2 + 2  shift upwards two units. g (x) = (x – 3)2 + 2  shift right three units. - 4 x y 4 f (x) = (x – 3)2 + 2 y = x 2 vertex (3, 2)

a > 0  parabola opens upward like y = 2x2. Shift left 1 and down 3 The standard form for the equation of a quadratic function is: f (x) = a(x – h)2 + k (a  0) The graph is a parabola opening upward if a  0 and opening downward if a  0. The axis of symmetry is x = h, and the vertex is (h, k). Example: Graph the parabola f (x) = 2x2 + 4x – 1 and find the axis of symmetry and vertex. x y f (x) = 2x2 + 4x – 1 x = –1 f (x) = 2x2 + 4x – 1 original equation f (x) = 2( x2 + 2x) – 1 factor out 2 f (x) = 2( x2 + 2x + 1) – 1 – 2 complete the square f (x) = 2( x + 1)2 – 3 standard form a > 0  parabola opens upward like y = 2x2. Shift left 1 and down 3 (–1, –3) axis x = –1, vertex (–1, –3).

a < 0  parabola opens downward. axis x = 3, vertex (3, 16). Example: Graph and find the vertex and x-intercepts of f (x) = –x2 + 6x + 7. x y 4 f (x) = – x2 + 6x + 7 original equation (3, 16) x = 3 f (x) = – ( x2 – 6x) + 7 factor out –1 f (x) = – ( x2 – 6x + 9) + 7 + 9 complete the square f (x) = – ( x – 3)2 + 16 standard form a < 0  parabola opens downward. axis x = 3, vertex (3, 16). Find the x-intercepts by solving –x2 + 6x + 7 = 0 (7, 0) (–1, 0) x2 – 6x – 7 = 0 divide by -1 (x – 7)( x + 1) = 0 factor x = 7, x = –1 x-intercepts (7, 0), (–1, 0) f(x) = –x2 + 6x + 7

Finding the Equation of a Parabola Find the equation for a parabola whose vertex is (1, 2) and that passes through the point (3, -2). Because the parabola has a vertex at (h, k)  (1, 2), the standard form equation is Now because the parabola passes through the point (3, -2), it follows that f(3) = -2

Now we can finish the standard form equation using a = -1 and Changing to quadratic form Quadratic form

Vertex of a Parabola The vertex of the graph of f (x) = ax2 + bx + c (a  0) Example: Find the vertex of the graph of f (x) = x2 – 10x + 22. f (x) = x2 – 10x + 22 original equation a = 1, b = –10, c = 22 At the vertex, So, the vertex is (5, -3).

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: 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.

Graphing Quadratic Functions Summary: Graph a quadratic function Write a quadratic function in standard form Find the equation of a parabola

Sneedlegrit: Find an equation of the parabola in standard form given the vertex at (5, 4) and that it goes through the point (6, 7). HW: page 214 (1-12, 27- 30, 37- 40, 47-50)