How do you write #f(x)=2x^2-7x-4# in vertex form?

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Answer 1 · 2016-08-14T04:07:01

#y=2(x-7/4)^2-81/8#

Given -

#y=2x^2-7x-4#

Find the vertex #(x, y)#

#x=(-b)/(2a)=(-(-7))/(2xx2)=7/4#

At #x=7/4#

#y=2(7/4)^2-7(7/4)-4=-81/8#

#x, y # coordinates of the vertex are #(7/4, -81/4)#

The vertex form of the quadratic equation is -

#y=a(x-h)^2+k#

We need the values of #a, h, k#

#a =2# [coefficient of #x^2#]
#h=7/4# [x-coordinate of the vertex]
#k=(-81)/8#[ y-coordinate of the vertex]

Now plug in the values in the equation

#y=2(x-7/4)^2-81/8#