Question

Find the x-intercepts of the parabola with vertex (-4,2) and y-intercept (0,-30). Write your answer in this form: (x1,y1),(y1,y2). If necessary, round to the nearest hundredth.

Answer 1

assuming that this parabola is inclined....etc....
so use the general equation
Ax^2 + 2B xy + Cy^2 + Dx + Ey + F = 0

as these points lie on the parabola..... put them and solve....

shortcut....
you need x intercept where y = 0

put y=0 in this eqn -
Ax^2 + 0 + 0 + Dx +0 + F = 0

Ax^2 + Dx + F = 0

Now eqn is reduced to x only, where again we have one point (0,-30)....here put x=0
A*0 + D*0 + F = 0 => F=0

eqn is now Ax^2 + Dx = 0

now with point (-4,2), x = -4

A (-4)^2 + D(-4) = 0
A*16-4D = 0
16A = 4 D
4A = D

or D = 4A

now put this in eqn Ax^2 + Dx = 0
=> Ax^2 + 4Ax = 0
=>x^2 + 4x =0
=>x(x+4) = 0

so x =0, x=-4 are intercepts on x axis for this parabola