4 Section 1.5: Quadratic Functions

a. The x-intercepts are at [latex](2,0)[/latex] and [latex](-4,0)[/latex]. So, plugging this into factored form we get [latex]f(x) = a(x-2)(x+4)[/latex]. Then we plug in [latex](0,16)[/latex] to solve for [latex]a[/latex]. So [latex]16 = a(0-2)(0+4)[/latex]. So, [latex]a = -2[/latex]. Thus, [latex]f(x) = -2(x-2)(x+4)[/latex]

a. Plugging the vertex into vertex form, we have [latex]y = a(x-1)^2-1[/latex]. Then we plug in the y-intercept [latex](0,6)[/latex] to solve for [latex]a[/latex], so [latex]6 = a(0-1)^2-1[/latex]. Solving for a we get [latex]a = 7[/latex]. Thus, [latex]y = 7(x-1)^2-1[/latex].

b. Plugging the vertex into factored form we have [latex]y = a(x-2)(x+3)[/latex]. Then we plug in the [latex]y[/latex]-intercept [latex](0,-3)[/latex] to solve for a so [latex]-3 = a(0-2)(0+3)[/latex]. Solving for [latex]a[/latex] we get [latex]a = \frac{1}{2}[/latex]. Thus, [latex]y = \frac{1}{2}(x-2)(x+3)[/latex].

c. Plugging the vertex into vertex form we have [latex]y = a(x+4)^2+3[/latex]. Then we plug in the point [latex](2,5)[/latex] to solve for [latex]a[/latex] so [latex]5 = a(2+4)^2 + 3[/latex]. Solving for [latex]a[/latex] we get [latex]a = \frac{1}{18}[/latex]. Thus, [latex]y = \frac{1}{18}(x+4)^2+3[/latex].

a. [latex]y = x^2-3x-18[/latex].

Vertex: [latex](1.5, -20.25)[/latex]

x-intercepts: [latex](6,0)[/latex] and [latex](-3,0)[/latex]

y-intercept: [latex](0,-18)[/latex]

b. [latex]y = 4x^2-4x-8[/latex]

Vertex: [latex](0.5,-9)[/latex]

x-intercepts: [latex](-1,0)[/latex] and [latex](2,0)[/latex]

y-intercept: [latex](0,-8)[/latex]

c. [latex]y = x^2+3x-21[/latex]

Vertex: [latex](-1.5, -23.25)[/latex]

x-intercepts: [latex](\frac{-3+\sqrt{93}}{2}, 0), (\frac{-3-\sqrt{93}}{2}, 0)[/latex]

y-intercept: [latex](0,-21)[/latex]

d. [latex]y = 2(x-7)^2+1[/latex]

Vertex: [latex](7, 1)[/latex]

x-intercepts: none

y-intercepts: [latex](0,99)[/latex]

a. The maximum height of the ball is 30.625 meters.

b. The ball hits the ground at [latex]t = 4[/latex] seconds.