Question

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

Answer 1

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.

Explanation:

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.