What is the vertex of #y= 2(x - 4)^2 - 8x+3 #?

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Answer 1 · 2017-04-26T16:53:07

The vertex is #(6,-27)#

Given: #y= 2(x - 4)^2 - 8x+3#

Expand the square:

#y= 2(x^2 - 8x+16) - 8x+3#

Distribute the 2:

#y= 2x^2 - 16x+32 - 8x+3#

Combine like terms:

#y= 2x^2 - 24x+35#

The x coordinate of the vertex, h, can be computed using the following equation:

#h = -b/(2a)# where #b = -24# and #a = 2#

#h = -(-24)/(2(2)#

#h = 6#

The y coordinate of the vertex, k, can be computed by evaluating the function at the value of h, (6):

#k= 2(6 - 4)^2 - 8(6)+3#

#k = -37#

The vertex is #(6,-27)#