How do you write an equation of the parabola that passes through the point (2, 5) and has x-intercepts -2 and 4?

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Answer 1 · 2016-11-20T12:22:05

There is a family of parabolas
#f(x,y;h)=(x+hy)^2-2x+(8/5-4h-5h^2)y-8=0#. The inserted graphs are for two members, with h = 0 and h = 1.

graph{x^2-2x+8/5y-8=0 [-20, 20, -10, 10]}

graph{x^2+2xy+y^2-2x-37/5y-8=0 [-20, 20, -10, 10]}

The general second degree equation represents a parabola, when

the second degree terms form a perfect square.

Let the equation be

#(x+hy)^2+ax+by+c=0#

The parabola passes through (-2, 0), (4, 0) and (2, 5). So,

#4-2a+c=0#,

#16+4a+c=0# and

#(2+5h)^2+2a+5b+c=0#

Solving this 3 X 4 system,

a = -2, c=-8 and b=8/5-4h-5h^2

See the answer.