How do you write the vertex form equation of the parabola # y=-x^2+4x+12#?

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Answer 1 · 2017-02-12T06:41:22

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

Vertex form of equation is of the type #y=a(x-h)^2+k#

Note that here coefficient of #x^2# is #a#, hence in #y=-x^2+4x+12#

we take #a=-1#. Then we go on to form a square of form #(x-h)^2#

#:.y=-x^2+4x+12#

= #-(x^2-2xx2xx x+2^2-2^2)+12#

#-((x-2)^2-4)+12#

= #-(x-2)^2+4+12#

= #-(x-2)^2+16#