Question

What is the standard form of `y= -5(x-8)^2 + 11`?

Answer 1

`y=-5x^2+80x-309`

Explanation:

The Standard Form for writing a polynomial is to put the terms with the highest degree first (what index it is raised to).

First, let's expand out the brackets:
`y= -5(x^2-8x-8x+64)+11`

`y=-5x^2+80x-320+11`

Simplify it, and make sure the terms are descending by their degree and you get

`y=-5x^2+80x-309`

Hope this helped; let me know if I can do anything else:)

Answer 2

`y=-5x^2+80x-309`

Explanation:

The standard form of the quadratic equation is

`y=ax^2+bx+c`

However, you have been given an equation in vertex form

`y=-5(x-8)^2+11`

First, factor out the `(x-8)^2` term using the FOIL process

`y=-5(x-8)(x-8)+11`
`y=-5(x^2-8x-8x+64)+11`
`y=-5(x^2-16x+64)+11`

Next, multiply the -5 through the factored expression.

`y=-5x^2+80x-320+11`

Finally, add the last two terms

`y=-5x^2+80x-309`