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#