Algebra 2 honors help please!?

Find b and c so that the parabola y=-3x^2+bx+c has x intercepts 1 and -4.


Please. I need all the help I can get with this. Thank you!!! ASAP!

For the parabola to have x-intercepts 1 and -4, y needs to be 0 when x is 1 and when x is -4.

So, first plug in 0 for y and 1 for x:
y = -3x² + bx + c
-3*1² + b*1 + c = 0
b + c - 3 = 0
b + c = 3

Now plug in 0 for y and -4 for x:
y = -3x² + bx + c
-3(-4)² + b(-4) + c = 0
c - 4b - 48 = 0
c - 4b = 48

Now we have a system of two equations:
b + c = 3
c - 4b = 48

Solve the first equation for c:
b + c = 3
c = 3 - b

Substitute that into the second equation and solve for b:
c - 4b = 48
3 - b - 4b = 48
3 - 5b = 48
5b + 48 = 3
5b = -45
b = -9

Plug that back into the equation c = 3 - b to find c:
c = 3 - b
c = 3 - (-9)
c = 12

So, the solution is b = -9, c = 12.

Hope that helps! :)