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

May 27, 2017

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

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:
$y = - 5 \left({x}^{2} - 8 x - 8 x + 64\right) + 11$

$y = - 5 {x}^{2} + 80 x - 320 + 11$

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

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

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

May 27, 2017

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

Explanation:

The standard form of the quadratic equation is

$y = a {x}^{2} + b x + c$

However, you have been given an equation in vertex form

$y = - 5 {\left(x - 8\right)}^{2} + 11$

First, factor out the ${\left(x - 8\right)}^{2}$ term using the FOIL process

$y = - 5 \left(x - 8\right) \left(x - 8\right) + 11$
$y = - 5 \left({x}^{2} - 8 x - 8 x + 64\right) + 11$
$y = - 5 \left({x}^{2} - 16 x + 64\right) + 11$

Next, multiply the -5 through the factored expression.

$y = - 5 {x}^{2} + 80 x - 320 + 11$

Finally, add the last two terms

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