What is the vertex form of #y= -6x^2 -27x-18#?

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Answer 1 · 2017-10-11T19:24:38

#y=-6(x+2.25)^2-109.5#

Currently your equation is in standard form:

#y=ax^2+bx+c# where #(-b/(2a),f(-b/(2a)))# is the vertex

We want to put it in vertex form:

#y=a(x-h)^2+k# where #(h,k)# is the vertex

We know #a=-6#, but we have to figure out the vertex to find #h# and #k#

#-b/(2a)=-(-27)/(2(-6))=(27/-12)=(-9/4)=-2.25#

So:

#f(-2.25)=-6(-2.25)^2-27(-2.25)-18#

#=-30.375-60.75-18=-109.5#

Thus our vertex is #(-2.25, -109.5)# and #h=-2.25, k=-109.5#

Thus our equation is:

#y=-6(x+2.25)^2-109.5#