Question

I need the help with question 1. Parts a b and c

Answer 1

From the problem, we have :

`f(x)=2(x-3)^2-6`

The vertex is (h, k) in the form :

`f(x)=a(x-h)^2+k`

a. So the vertex will be (3, -6)

b. y-intercept is the value of f(x) when x = 0.

`\begin{gathered} f(0)=2(0-3)^2-6_{} \\ =2(9)-6 \\ =12 \end{gathered}`

y-intercept = (0, 12)

c. x-intercept is the value of x when f(x) = 0.

`\begin{gathered} 0=2(x-3)^2-6 \\ -2(x-3)^2=-6 \\ (x-3)^2=(-6)/(-2) \\ (x-3)^2=3 \\ \sqrt[]{(x-3)^2}=\sqrt[]{3} \\ x-3=\pm\sqrt[]{3} \\ x=3\pm\sqrt[]{3} \end{gathered}`

The x-intercepts are :

`(3+\sqrt[]{3},0)\quad \text{and}\quad (3-\sqrt[]{3},0)`