Question

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

Answer 1

The vertex `(h,k)=(1,8)`
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:

`-2x^2+4x+6=0`

To solve it you you factor it out:

`-2(x^2-2x-3)`
`-2(x-3)(x+1)`

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(0)^2+4(0)+6`

`y=6`

So, the y-intercept is `6`

To find the vertex `(h,k)` you use your equation to plug into the general formula.

Remember that `ax^2+bx+c`

To find `h`:

`h=-b/(2a)`

`h=-4/((2)(-2))=1`

`h=1`

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

`k=-2(1)^2+4(1)+6`

`k=8`

The vertex `(h,k)=(1,8)`

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]}