How do you write #f(x)= -2x^2 + 5x - 8# in vertex form?

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Answer 1 · 2016-09-07T05:59:39

#y=-2(x-5/4)^2-39/8#

Given -

#y=-2x^2+5x-8#

X-co-ordinate of the vertex

#x=(-b)/(2a)=(-(5))/(2 xx (-2))=(-5)/(-4)=5/4#
Y-coordinate of the vertex
At #x=5/4#

#y=-2(5/4)^2+5(5/4)-8=-39/8#

Vertex form of the equation is

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

#a=-2# coefficient of #x^2#
#h=5/4# #x#coordinate of the vertex
#k=-39/8# #y#coordinate of the vertex

Equation-

#y=-2(x-5/4)^2-39/8#