Find b and c of the equation

Question

Find b and c so that y= -12x^2 +bx+c has the vertex (-5,-8)

Answer 1

Hey! Lets solve this shall we ?

So, we must find out what b and c is with the given information, right ? First, we must yield the Vertex Formula such that

V = (-b/2a, f(-b/2a) = (-5, -8).........Then we let the x value equal -5 such that

-b / 2a = -5............Since a = -12, plug that in to get

-b / -24 = -5...........We cross multiply to obtain

-b = 120.........Then divide -1 to get

b = -120 ! ..........Then you pull out the Quadratic Formula to find c !

I hope this helped!

Answer 2

f(x) = -12(x + 5)² - 8

f(x) = -12(x² + 10x + 25) - 8

f(x) = -12x² - 120x - 300 - 8

f(x) = -12x² - 120x - 308

Answer 3

For y = Ax² + Bx + C, the x-coordinate of the vertex is -B/(2A).

So, if y = -12x² + bx + c, then the x-coordinate of the vertex is b / 24 = -5. Thus, b = -120.

So, y = -12x² - 120x + c

Since the vertex is a point on the graph, we have -8 = -12(-5)² -120(-5) + c. So, c = -308.

Answer 4

The "x" coordinate of the vertex of a quar=dratic can be determined by x = -b/(2a) so since a = -12 and the x coordinate of the vertex is -5, we can say:

-5 = -b/[(2)(-12)]

-5 = -b/(-24) [multiply both sides by -24 to get:

-b = 120

b = -120

The "y" coordinate of the vertex can be determined by plugging in the x-coordinate so, since the y-coordinate of the vertex is -8, we can say:

-8 = -12(-5)² - 120(-5) + c

-8 = -300 + 600 + c

-8 = 300 + c

c =-308