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

Nov 15, 2015

$x = 6$,
$x = 0$.

#### Explanation:

$y = 2 {x}^{2} - 12 x$.

In order to find the $y$-intercept, we must substitute $x$ by 0, as the $y$-axis is $x = 0$. So,
$y = 2 {\left(0\right)}^{2} - 12 \left(0\right)$,
$y = 0 - 0$,
$y = 0$.

In order to find the $x$-intercept, we must substitute $y$ by 0, as the $x$-axis is $y = 0$. So,
$0 = 2 {x}^{2} - 12 x$,
$2 \left({x}^{2} - 6 x\right) = 0$,
${x}^{2} - 6 x = 0$, according to $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$,
$x = \frac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{2} - 4 \left(1\right) \left(0\right)}}{2 \left(1\right)}$,
$x = \frac{6 \pm \sqrt{36}}{2}$,
$x = \frac{6 + 6}{2}$,
$x = \frac{12}{2}$,
$x = 6$,
$x = \frac{6 - 6}{2}$,
$x = \frac{0}{2}$,
$x = 0$.

Hope it Helps! :D .