# How do you find the vertex, x and y intercepts for y=-2x^2+4x+6?

Jun 16, 2017

The vertex $\left(h , k\right) = \left(1 , 8\right)$
x-intercepts are $x = - 1 , 3$
The y-intercept is $y = 6$

#### Explanation:

To find the x intercepts you set the equation equal to zero and solve:

$- 2 {x}^{2} + 4 x + 6 = 0$

To solve it you you factor it out:

$- 2 \left({x}^{2} - 2 x - 3\right)$
$- 2 \left(x - 3\right) \left(x + 1\right)$

Now set both factors equal to zero and solve:

$x - 3 = 0$
$x = 3$

$x + 1 = 0$
$x = - 1$

So, the x-intercepts are $x = - 1 , 3$

To find the y-intercept its a bit easier you simply set the x-values equal to zero:

$y = - 2 {\left(0\right)}^{2} + 4 \left(0\right) + 6$

$y = 6$

So, the y-intercept is $6$

To find the vertex $\left(h , k\right)$ you use your equation to plug into the general formula.

Remember that $a {x}^{2} + b x + c$

To find $h$:

$h = - \frac{b}{2 a}$

$h = - \frac{4}{\left(2\right) \left(- 2\right)} = 1$

$h = 1$

Now to find $k$ we plug in our $h$ value of $1$ into $x$:

$k = - 2 {\left(1\right)}^{2} + 4 \left(1\right) + 6$

$k = 8$

The vertex $\left(h , k\right) = \left(1 , 8\right)$

As you can see by the graph below, we got the right answers.

graph{-2x^2+4x+6 [-8.68, 11.32, -0.485, 9.515]}