Question

How do you find the vertex and the intercepts for `f(x) = -2(x+2)^2`?

Answer 1

In vertex form, the vertex is given as the `h` and `k` values.

The intercepts, just sub in each variable as `0` and solve for the other.

Explanation:

The equation is given to us in vertex form, making the process of determining the vertex much easier.

The base formula for vertex form is `f(x) = a(x-h)+k`
=> Where `h` is the x-coordinate of the vertex.
=> Where `k` is the y-coordinate of the vertex.
=> Where `a` is factor of how much the parabola stretches/compresses by in a vertical sense.

The "`(x+2)`" gives us the x-coordinate of the vertex as `x=-2`. There is no `k`-value, meaning the y-coordinate is on the x-axis.

Therefore, the vertex of the equation is `(-2,0)`.

The intercepts can be determined by simply subbing in each of the variables as 0, and solving for the other.

First, we'll find the x-intercept. Therefore, we'll sub in the y-value as `0`.

`f(x) = -2(x+2)^2`

`y = -2(x+2)^2`

I switched `f(x)` into `y` to make comprehension easier

`0 = -2(x+2)^2`

`0 = (x+2)^2`

`0 = x+2`

`-2 = x`

And now the y-intercept. Therefore, we'll sub in the x-value as `0`.

`f(x) = -2(x+2)^2`

`y = -2(x+2)^2`

I switched `f(x)` into `y` to make comprehension easier

`= -2(0+2)^2`

`= -2(2)^2`

`= -2(4)`

`= -8`

Your parabola will look like this.

`f(x) = -2(x+2)^2` graph{-2(x+2)^2 [-10.875, 9.125, -6.28, 3.72]}

As you can see, the vertex is in fact `(-2, 0)`, and the intercepts are `(-2,0)` and `(0,-8)`.

Hope this helps :)