# How do you find the intercept and vertex of y = 2x^2 + 8x − 4?

Mar 24, 2017

Vertex: $\left(- 2 , - 12\right)$
y-intercept: $\left(0 , - 4\right)$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} y = 2 {x}^{2} + 8 x - 4$

Our initial objective will be to convert this into "vertex-form",
namely $y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Setting aside the constant $\left(- 4\right)$ for a moment and extracting the $\textcolor{g r e e n}{m}$ factor:
$\textcolor{w h i t e}{\text{XXX")y=color(green)2(x^2+4x)color(white)("XXX}} - 4$

Completing the square:
$\textcolor{w h i t e}{\text{XXX")y=color(green)2(x^2+4xcolor(magenta)(+4))color(white)("XX}} - 4 \textcolor{m a \ge n t a}{- \left(\textcolor{g r e e n}{2} \times 4\right)}$

Rewriting as a squared binomial and a simplified constant:
$\textcolor{w h i t e}{\text{XXX")y=color(green)2(x+2)^2color(white)("XX}} - 12$

Express in explicit vertex form:
$\textcolor{w h i t e}{\text{XXX")y=color(green)2(x-color(red)(""(-2)))^2+color(blue)(} \left(- 12\right)}$

color(orange)("Vertex is at: "(color(red)(-2),color(blue)(-12))

The y-intercept is the value of $y$ when $x = 0$.
Substituting $0$ for $x$ in the original equation :
$\textcolor{w h i t e}{\text{XXX}} y = 2 \cdot {0}^{2} - 8 \cdot 0 - 4$
gives the color(cyan)("y-intercept at "y=-4 or, if you prefer at color(cyan)(""(0,-4))

graph{2x^2+8x-4 [-12.78, 12.53, -12.6, 0.06]}