Which of the following is the vertex of the quadratic equation y = 6x^2 - 24x + 32?

(-2, -8)
(-2, 8)
(2, -8)
(2, 8)

Question

Which of the following is the vertex of the quadratic equation y = 6x^2 - 24x + 32?
 	
	(-2, -8)
	(-2, 8)
	(2, -8)
	(2, 8)

anonymous · Answer

so i know the vertex form is y=a(x-h)^2 +k

anonymous · Answer

idk how to turn y=6x^2 - 24 + 32 into the vertex form

driftracer305 · Answer

factorize it....

campbell_st · Answer

so start this way 

y = 6(x^2 - 4x) + 32

what needs to be added to get the perfect square...?

anonymous · Answer

i dunno

campbell_st · Answer

well the perfect square would be 

y = 6(x^2 - 4x + 4) + 32

so how do you compensate..?

campbell_st · Answer

or what would I have added to create the perfect square in the brackets..?

anonymous · Answer

+4 ?

campbell_st · Answer

yep... but because the common factor of 6 is outside I really needed to add 24

6 x 4 = 24... 

so compensate I now subract 24 from 32

so you have 

$$y = 6(x^2 -4x + 4) + 32 - 24$$

wich can be simplified to 

$$y = 6(x -2)^2 + 8$$

so any ideas on the vertex...?

anonymous · Answer

(2,8)

campbell_st · Answer

yep... looks good to me.... to check you can subsitute x = 2 into the equation and you should get y = 2

campbell_st · Answer

oops y = 8