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

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Answer 1 · 2017-05-27T19:39:14

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

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 · 2017-05-27T19:39:44

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

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

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

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