How do you find the vertex and intercepts for y = 1/20x^2?

Jun 29, 2016

Vertex is at $\left(0 , 0\right)$
y-intercept at $0$
x-intercept (only one) at $0$

Explanation:

One way to look at this is to considered the standard vertex form:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

the given equation: $y = \frac{1}{20} {x}^{2}$ can be written in explicit standard vertex form as:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{\frac{1}{20}} {\left(x - \textcolor{red}{0}\right)}^{2} + \textcolor{b l u e}{0}$
with vertex at $\left(\textcolor{red}{0} , \textcolor{b l u e}{0}\right)$

The y-intercept is the value of $y$ when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} y = \frac{1}{20} {\left(0\right)}^{2} = 0$

The x-intercept(s) is(are) the value(s) of $x$ when $y = 0$
$\textcolor{w h i t e}{\text{XXX}} 0 = \frac{1}{20} {x}^{2} \rightarrow x = 0$