What are the vertex, focus and directrix of # y=2x^2 +11x-6 #?

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Answer 1 · 2016-12-19T07:51:01

The vertex is #=(-11/4,-169/8)#
The focus is #=(-11/4,-168/8)#
The directrix is #y=-170/8#

Let rewrite the equation

#y=2x^2+11x-6#

#=2(x^2+11/2x)-6#

#=2(x^2+11/2x+121/16)-6-121/8#

#y=2(x+11/4)^2-169/8#

#y+169/8=2(x+11/4)^2#

#1/2(y+169/8)=(x+11/4)^2#

This is the equation of the parabola

#(x-a)^2=2p(y-b)#

The vertex is #=(a,b)=(-11/4,-169/8)#

The focus is #=(a,b+p/2)=(-11/4,-169/8+1/8)#

#=(-11/4,-168/8)#

The directrix is #y=b-p/2#

#=>#, #y=-169/8-1/8=-170/8#

graph{(y-2x^2-11x+6)(y+170/8)=0 [-14.77, 10.54, -21.49, -8.83]}