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

2 Answers
May 27, 2017

#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:)

May 27, 2017

#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#