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

Dec 4, 2015

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

#### Explanation:

vertex form: $y = a {\left(x - h\right)}^{2} + k$, with vertex $\left(h , k\right)$

vertex $= \left(3 , 1\right)$

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 $x = 0$

$y \left(0\right) = 2 \left({3}^{2}\right) + 1 = 19$

hope that helped