On a parabla if the line crosses x at -1 and -5 , how do find the quadratic function?

Answer 1

The general equation for a parabola is

(y - k)² = 4a(x - h)

where h and k are the x- and y-coordinates of the vertex and the focus is at ( h, k + a ).

You have provided two points on the parabola: ( -1, 0 ) and ( -5, 0 ). Alas, you have three unknowns. However, you can plug in the points and solve for h and k in terms of a.

Answer 2

(x +5)^3(x + 1)

or (x + 5)(x + 1)^3

Answer 3

the standard form for intercept is y=a(x-p)(x-q)
where it crosses would be -1 and -5 so p=-1, q =-5
after that find the middle/between point of it which is -3 so plug it in which turns to y=(-3+1)(-3+5) leave out a because theres nothing for a,
add and multiply that together to get y=-4
so vertex is -3,-4
the parabola should open up

Answer 4

This is telling you the two roots of your equation, which are -1 and -5.
(x+5)(x+1)=0
expand gives
x^2+x+5x+5
x^2+6x+5
this is the equation you want
(note that ^ means x-squared)