Question

What is the vertex of `y= -x^2+40x-16`?

Answer 1

The vertex is at `(20, 384)`.

Explanation:

Given: `y = -x^2 + 40x - 16`

This equation is in standard quadratic form `(y = ax^2 + bx + c)`, meaning we can find the `x`-value of the vertex using the formula `(-b)/(2a)`.

We know that `a = -1`, `b = 4`, and `c = -16`, so let's plug them into the formula:
`x = (-40)/(2(-1)) = 20`

Therefore, the `x`-coordinate is `20`.

To find the `y`-coordinate of the vertex, plug in the `x`-coordinate and find `y`:
`y = -x^2 + 40x - 16`

`y = -(20)^2 + 40(20) - 16`

`y = -400 + 800 - 16`

`y = 384`

Therefore, the vertex is at `(20, 384)`.

Hope this helps!