# How do you find the vertex and intercepts for y = 2x^2?

Nov 14, 2015

They are all at $\left(0 , 0\right)$.

#### Explanation:

Finding the vertex:
Use the formula $\left(- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right)$, given the form of the parabola $f \left(x\right) = a {x}^{2} + b x + c$. In your case, $a = 2 , b = 0$ and $c = 0$.
Here, $- \frac{b}{2 a} = - \frac{0}{2 \cdot 2} = 0$ and $f \left(- \frac{b}{2 a}\right) = f \left(0\right) = 2 {\left(0\right)}^{2} = 0$.
Thus, the vertex is located at $\left(0 , 0\right)$.

Finding the X-intercept:
Plug in $0$ for $y$.
$0 = 2 {x}^{2} \implies 0 = {x}^{2} \implies 0 = x$
Then plug in $0$ for $y$ to determine the point of the intercept. We already know that if we plug in $0$ for either point, we will receive $0$ as our answer, so the X-intercept is $\left(0 , 0\right)$.

Finding the Y-intercept:
This is very similar to finding the X-intercept, in that we now plug in $0$ for $x$, which we already know will result in the point $\left(0 , 0\right)$.

Alternative method: Graph the parabola and look at the vertex and intercepts.