From the problem, we have :
[tex]f(x)=2(x-3)^2-6[/tex]
The vertex is (h, k) in the form :
[tex]f(x)=a(x-h)^2+k[/tex]
a. So the vertex will be (3, -6)
b. y-intercept is the value of f(x) when x = 0.
[tex]\begin{gathered} f(0)=2(0-3)^2-6_{} \\ =2(9)-6 \\ =12 \end{gathered}[/tex]
y-intercept = (0, 12)
c. x-intercept is the value of x when f(x) = 0.
[tex]\begin{gathered} 0=2(x-3)^2-6 \\ -2(x-3)^2=-6 \\ (x-3)^2=\frac{-6}{-2} \\ (x-3)^2=3 \\ \sqrt[]{(x-3)^2}=\sqrt[]{3} \\ x-3=\pm\sqrt[]{3} \\ x=3\pm\sqrt[]{3} \end{gathered}[/tex]
The x-intercepts are :
[tex](3+\sqrt[]{3},0)\quad \text{and}\quad (3-\sqrt[]{3},0)[/tex]
