# What are the y and x intercept(s) of y=2x^2-4?

Sep 29, 2016

We can set alternately $x = 0$ and $y = 0$ to find the intercepts:

#### Explanation:

To find the y-intercept set $x = 0$ into your expression and get:
$y = 2 \cdot 0 - 4 = - 4$
Sothe coordinates of the y-intercept will be:
$x = 0 \mathmr{and} y = - 4$

To find the x-intercept(s) set $y = 0$ to get:
$2 {x}^{2} - 4 = 0$
Rearranging:
${x}^{2} = \frac{4}{2}$
${x}^{2} = 2$
$x = \pm \sqrt{2}$
We have two intercepts of coordinates:
$x = \sqrt{2} \mathmr{and} y = 0$
$x = - \sqrt{2} \mathmr{and} y = 0$

Graphically we can "see" them:
graph{2x^2-4 [-8.625, 11.375, -6.64, 3.36]}

Sep 29, 2016

y-intercept: $y = - 4$
x-intercepts: $x = - \sqrt{2} \mathmr{and} x = \sqrt{2}$

#### Explanation:

The y-intercept is the value of $y$ when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} y = 2 {x}^{2} - 4$ with $x = 0$ becomes
$\textcolor{w h i t e}{\text{XXX}} y = 2 \cdot {0}^{2} - 4 = - 4$

The x-intercepts are the values of $x$ when $y = 0$
$\textcolor{w h i t e}{\text{XXX}} y = 2 {x}^{2} - 4$ when $y = 0$ becomes
$\textcolor{w h i t e}{\text{XXX}} 0 = 2 {x}^{2} - 4$
$\textcolor{w h i t e}{\text{XXX}} 2 {x}^{2} = 4$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} = 2$
$\textcolor{w h i t e}{\text{XXX}} x = {+}_{\sqrt{2}}$

Sep 29, 2016

$y$ intercept $- 4$, $x$ intercepts $\pm \sqrt{2}$

#### Explanation:

$y = 2 {x}^{2} - 4$

The $y$ intercept is at $x = 0$
Where: $y = - 4$

The $x$ intercept ia at $y = 0$
Where: $2 {x}^{2} - 4 = 0$

${x}^{2} = \frac{4}{2}$

$x = \pm \sqrt{2}$

These can be seen on the graph of $2 {x}^{2} - 4$ below
graph{2x^2-4 [-6.1, 6.384, -5.12, 1.126]}