Question

How do you find the vertex and intercepts for `y = 2(x - 3)^2 + 1`?

Answer 1

This is already in vertex form making it easy to identify the vertex ...

Explanation:

vertex form: `y=a(x-h)^2+k`, with vertex `(h,k)`

vertex `=(3,1)`

The vertex is above the x-axis and since the coefficient `a=2` is positive, this parabola opens upward and will have no x-intercepts (only 2 imaginary roots).

The y-intercept happens when `x=0`

`y(0)=2(3^2)+1=19`

hope that helped