Determine the roots (or zeros) of the function: $y=2(x+3)^2-32$ ?

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Answer 1 · 2018-02-21T17:15:28

See below:

I'm assuming the equation is $y=2(x+3)^2-32$

To find the zeroes, set $y$ as $0$ .

$0=2(x+3)^2-32$

Solve:

$0=2(x^2+6x+9)-32$

$0=2x^2+12x+18-32$

$0=2x^2+12x-14$

$0=2(x^2+6x-7)$

Factor:

$0=2(x+7)(x-1)$

$x=-7,1$

Determine the roots (or zeros) of the function: y=2(x+3)^2-32y=2(x+3)^2-32 ?