Quadratic Functions and Their Properties Section 4.3 Quadratic Functions and Their Properties

R(p) = (21,000 – 150p)p = -150p2 +21,000p

OBJECTIVE 1

vertex: (-2, -3) axis of symmetry: x = -2

OBJECTIVE 2

Without graphing, locate the vertex and axis of symmetry of the parabola defined by . Does it open up or down? Vertex: (1,4) Axis of symmetry: x = 1 Opens Down

OBJECTIVE 3

Vertex: (-2,-17) x-intercepts: (-4.38, 0), (0.38, 0) y-intercept: (0, -5) Domain: All Reals, Range y -17 Decreasing: (-∞,-2), Increasing: (-2,∞)

Vertex: (-2,0) x-intercepts: (-2, 0) y-intercept: (0, 4) Domain: All Reals, Range y ≥ 0 Decreasing: (-∞,-2), Increasing: (-2,∞)

vertex: (2, 11) x-intercepts: (-1.32, 0), (5.32 , 0) y-intercept: (0, 7)

Domain: All Reals, Range y ≤ 11 Increasing: (-∞, 2), Decreasing: (2,∞)

Determine the quadratic function whose vertex is (-2, -5) and whose y intercept is – 1. y = a(x - h)2 + k where (h, k) is the vertex y = a[x – (-2)]2 + (-5) = a(x + 2)2 – 5 (0, -1): -1 = a(0 + 2)2 -5 -1 = 4a – 5 4 = 4a  a = 1 y = 1(x + 2)2 – 5 = x2 + 4x + 4 – 5 = x2 + 4x - 1

OBJECTIVE 4

Since a = -1 it will have a maximum at x = So the maximum value is