Quadratic Equation: a function that can be written in the form: f(x) = ax 2 + bx + c f(x) = x 2 Vertex: the graph’s “turning point.” Minimum or maximum point? Axis of symmetry – the line that divides the parabola into two halves that are mirror images of each other x = 0 xf(x) f(x) decreases on the x-interval (- , 0) f(x) increases on the x- interval (0,  )

f(x) = a(x – h) 2 + k where (h, k) is the vertex if a > 0, the graph opens upward (minimum) if a < 0, the graph opens downward (maximum)

Graph f(x) = -3(x + 2) 2 – 4 the vertex is at (-2, -4) for accuracy, pick a point to the left and a point to the right of the vertex – I usually pick the same distance. Shifts 2 left Shifts 4 down the graph is flipped (max) the graph is stretched (-2, -4) f(-3) = -3(-3 + 2) 2 – 4 = -7 f(-1) = -3(-1 + 2) 2 – 4 = -7 FYI: the axis of symmetry is x = -2

a) Find the vertex: f(x) = x 2 – 8x + 16 vertex form: f(x) = (x – 4) 2  vertex (4, 0) b) Find the vertex: f(x) = x 2 – 10x + 16 f(x) = x 2 – 10x + 25 – 9 f(x) = (x – 5) 2 – 9  vertex (5, -9)

If f(x) = (x – 9)(x +2) a) Find the x-intercepts (set f(x) = 0) 0 = (x – 9)(x + 2) (x – 9) = 0 (x + 2) = 0 x = 9 x = -2 (9, 0)(-2, 0)

x-intercepts: (9, 0), (-2, 0) b) Use the x-intercepts to find the vertex. find the x-coord of the vertex … x 1 + x 2 2 = 3.5 find the y-coord of the vertex … If f(x) = (x – 9)(x + 2) f(7/2) = (7/2 – 9)(7/2 + 2) = Vertex (3.5, 54.25)

If f(x) = ax 2 + bx + c, then the x-coord of the vertex = how do you find the y-coordinate?

Graph f(x) = -2x 2 + 5x +7 Find the vertex (blank space) Find 2 other points (blank space)