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

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Answer 1 · 2018-07-21T23:29:25

$y = 2 x^{2} + 16 x + 11$ $y = 2x^2 + 16x + 11$

Standard quadratic form is $y = a x^{2} + b x + c$ $y = ax^2 + bx + c$ .

$y = 2 {(x + 4)}^{2} - 21$ $y = 2(x+4)^2 - 21$

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

$y = 2 (x^{2} + 8 x + 16) - 21$ $y = 2(x^2 + 8x + 16) - 21$

$y = 2 x^{2} + 16 x + 32 - 21$ $y = 2x^2 + 16x + 32 - 21$

$y = 2 x^{2} + 16 x + 11$ $y = 2x^2 + 16x + 11$

As you can see, this is now in the form $y = a x^{2} + b x + c$ $y = ax^2 + bx + c$ .

Hope this helps!

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