What is the equation of the parabola with axis intercepts of #x=2#, #x=-4#, and #y=4#?

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Answer 1 · 2018-05-22T07:18:11

#y = -1/2x^2-x+4#

The #x# intercepts (#x=2# and #x=-4#) are the zeros of the quadratic function, corresponding to linear factors #(x-2)# and #(x+4)#.

So our equation can be written something like:

#y = a(x-2)(x+4)#

for some constant #a# to be determined.

Putting #x=0# and #y=4#, we have:

#color(blue)(4) = a(color(blue)(0)-2)(color(blue)(0)+4) = -8a#

Hence #a=-1/2# and we have:

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