# How do you find the vertex and the intercepts for y= 3x^2 +12x-6?

Jan 12, 2018

Vertex: $\left(- 2 , - 18\right)$
y-intercept: $\left(0 , - 6\right)$
x-intercepts: $\left(- 2 - \sqrt{6} , 0\right)$ and $\left(- 2 + \sqrt{6} , 0\right)$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} y = 3 {x}^{2} + 12 x - 6$

This can be converted into vertex form as follows:
$\Rightarrow y = 3 \left({x}^{2} + 4 x - 2\right)$

$\Rightarrow y = 3 \left[{x}^{2} + 4 x - 2\right]$

$\Rightarrow y = 3 \left[\left({x}^{2} + 4 x \textcolor{b l u e}{+ 4}\right) - 2 \textcolor{b l u e}{- 4}\right]$

$\Rightarrow y = 3 \left[{\left(x + 2\right)}^{2} - 6\right]$

$\Rightarrow y = 3 {\left(x + 2\right)}^{2} - 18$

rArr y= 3(x-color(red)(""(-2)))^2+color(blue)(""(-18))

which is the vertex form with vertex at $\left(\textcolor{red}{- 2} , \textcolor{b l u e}{- 18}\right)$

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The y-intercept is the value of $y$ when $x = \textcolor{m a \ge n t a}{0}$.
Substituting $\textcolor{m a \ge n t a}{0}$ for $x$ in the original equation
$\textcolor{w h i t e}{\text{XXX}} y = 3 \cdot {\textcolor{m a \ge n t a}{0}}^{2} + 12 \cdot \textcolor{m a \ge n t a}{0} - 6 = - 6$
So the y-intercept is at $y = - 6$ [or if you prefer at $\left(0 , - 6\right)$]

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The x-intercepts are the values of $x$ for points when $y = 0$

Using the form derived while determining the vertex form: $y = 3 {\left(x + 2\right)}^{2} - 18$
with $y = \textcolor{g r e e n}{0}$
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{0} = 3 {\left(x + 2\right)}^{2} - 18$

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow \textcolor{g r e e n}{0} = {\left(x + 2\right)}^{2} - 6$

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow {\left(x + 2\right)}^{2} = 6$

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow x + 2 = \pm \sqrt{6}$

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow x = - 2 \pm \sqrt{6}$

So the x-intercepts are at $x = - 2 - \sqrt{6}$ and $x = - 2 + \sqrt{6}$ [or if you prefer at $\left(- 2 - \sqrt{6} , 0\right)$ and $\left(- 2 + \sqrt{6} , 0\right)$]