What is the equation of the parabola that has a vertex at # (0, 8) # and passes through point # (5,-4) #?

1 Answer
Alan P.
Jan 22, 2016

There are an infinite number of parabolic equations that meet the given requirements.
If we restrict the parabola to having a vertical axis of symmetry, then:
#color(white)("XXX")y=-12/25x^2+8#

Explanation:

For a parabola with a vertical axis of symmetry, the general form of the parabolic equation with vertex at #(a,b)# is:
#color(white)("XXX")y=m(x-a)^2+b#

Substituting the given vertex values #(0,8)# for #(a,b)# gives
#color(white)("XXX")y=m(x-0)^2+8#

and if #(5,-4)# is a solution to this equation, then
#color(white)("XXX")-4=m((-5)^2-0)+8 rArr m=-12/25#

and the parabolic equation is
#color(white)("XXX")color(black)(y=-12/25x^2+8)#
graph{y=-12/25*x^2+8 [-14.21, 14.26, -5.61, 8.63]}

However, (for example) with a horizontal axis of symmetry:
#color(white)("XXX")color(black)(x=5/144(y-8)^2)#
also satisfies the given conditions:
graph{x=5/144(y-8)^2 [-17.96, 39.76, -8.1, 20.78]}

Any other choice for the slope of the axis of symmetry will give you another equation.

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