What is the standard form of #y= 2(x + 4)^2 - 21 #?

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Answer 1 · 2018-07-21T23:29:25

#y = 2x^2 + 16x + 11#

Standard quadratic form is #y = ax^2 + bx + c#.

#y = 2(x+4)^2 - 21#

First, simplify the expression in the parenthesis with exponent:
#y = 2(x+4)(x+4)-21#

#y = 2(x^2 + 8x + 16) - 21#

#y = 2x^2 + 16x + 32 - 21#

#y = 2x^2 + 16x + 11#

As you can see, this is now in the form #y = ax^2 + bx + c#.

Hope this helps!