Question

How do you find the vertex and the intercepts for `f(x) = -(x+5)^2`?

Answer 1

The vertex form for the equation of a parabola such that `y = f(x)` is:
`y = a(x - h)^2 + k" [1]"`

where `(h,k)` is the vertex.

Explanation:

Write the given equation in the form of equation [1]:

`y = -(x - -5)^2" [2]"`

The vertex is the point `(-5, 0)`

To find the x intercept(s), set equation [2] equal to zero and solve for x:

`0 = -(x - -5)^2`

`x = -5`

Because the vertex is, also, the x intercept, there is only one x intercept at the point `(-5, 0)`.

To find the y intercept set x, in equation [2], equal to zero and solve for y:

`y = -(-5)^2`

`y = -25`

The y intercept is the point `(0, -25)`