Question

How do you find the vertex of `F(x)=x^2+2x-8`?

Answer 1

-1

Explanation:

For a quadratic function like this, the vertices are points of symmetry.

We can use a shortcut such that:
`x=(-b)/(2a)`

In your case, `b` is 2 and `a` is 1. So:
`-2/(2*1)=-1`

Confirm it on a graph.
graph{x^2+2x-8 [-25.8, 25.8, -18.1, 7.7]}

Answer 2

The vertex is when the gradient is 0.

Explanation:

Differentiate and find `x` where `F'(x) = 0`

`F(x) = x^2 + 2x - 8`

`F'(x) = 2x + 2`

`0 = 2x+ 2`

`x = -1`

`F(-1) = (-1)^2 + 2(-1) - 8`

`F(-1) = -9`

Therefore the vertex is: `(-1, -9)`

Answer 3

Vertex is (-1, -9)

Explanation:

To find the x-value of the vertex, we use the formula `x=(-b)/(2a)` (This is considered the axis of symmetry).

`a=1` and `b=2` with `c=-8` (These are the numbers in order)

By substituting we get `x=(-(2))/(2*1)`, which simplifies to `x=-1`.

Once you find x, substitute back into the equation to find y.
`y=(-1)^2+2(-1)-8`
`y=1-2-8`
`y=-9`

Putting the two numbers together you get the coordinates `(-1, -9)`.