# How do you find the vertex and intercepts for y=x^2+6?

Jan 30, 2016

vertex at $\left(0 , 6\right)$
y-intercept at $6$
there is no x-intercept

#### Explanation:

vertex
The general vertex form is
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Re-writing the given: $y = {x}^{2} + 6$ into explicit vertex form:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{1} {\left(x - \textcolor{red}{0}\right)}^{2} + \textcolor{b l u e}{6}$
with vertex at ((color(red)(0),color(blue)(6))

y-intercept
The y-intercept is the value of $y$ when $x = 0$
Setting $x = 0$ in the given equation:
$\textcolor{w h i t e}{\text{XXX}} y = {\left(0\right)}^{2} + 6 = 6$
The y-intercept is at $y = 6$

x-intercept
The x-intercept is the value of $x$ which causes $y = 0$
For the given equation $y = {x}^{2} + 6$
$\textcolor{w h i t e}{\text{XXX}}$if $y = 0$ then ${x}^{2} = - 6$
$\textcolor{w h i t e}{\text{XXX}}$ but $x \ge 0$ for all Real values of $x$
Therefore
$\textcolor{w h i t e}{\text{XXX}}$there is no x-intercept.
graph{x^2+6 [-15.54, 12.94, -1.06, 13.18]}