The general form for the equation of a parabola is a quadratic: y=ax²+bx+c where a, b, c are constants. When y=0, x=-6 and -4. 0=36a-6b+c=16a-4b+c. So 36a-16a-6b+4b=0, 20a-2b=0, which reduces to 10a-b=0 so b=10a.
Now plug in the third point: -18=9a-3b+c. So c=-18-9a+3b. We can eliminate c by using one of the other equations:
Use c=4b-16a=-18-9a+3b, 4b-3b-16a+9a=-18, b-7a=-18. But b=10a, so 10a-7a=-18, 3a=-18, a=-6.
Knowing a we can find b=10a=-60, and, finally, c=4b-16a=-240+96=-144.
y=-6x²-60x-144=-6(x²+10x+24)=-6(x+6)(x+4).
Another way to do this is to note the x intercepts at -6 and -4. This means that x+6 and x+4 must be roots of the quadratic: y=a(x+6)(x+4). To find a we plug in the third point: -18=a(-3+6)(-3+4)=3a, making a=-6. This is probably the method you are expected to use, because the x intercepts are a giveaway!