Find b and c so that y = 16 x 2 + b x + c has vertex ( − 5 , − 6 )

Answer 1

16x² + bx + c = 16(x + 5)² - 6

Expand and determine values of b and c.

Answer 2

Hi Madison W.

b = 160

c = 394

 

In the problem we are given the x, y coordinates of the vertex (-5,-6) and the Standard Form of the parabola 16x² + bx + c.

Since standard form is y = ax² + bx + c

In y = 16x² + bx + c

a = 16

The x coordinate of the vertex of a parabola is (-b/2a) we can solve for b

-5 = -b/2(16)

-5 = -b/32

-5(32) = -b

-160 = -b

Divide both sides by -1

160 = b

We already know a = 16 and we have the y coordinate of the vertex, -6 we can calculate c

16x² + 160x + c = y

16(-5²) + 160(-5) + c = -6

16(25) - 800 + c = -6

400 - 800 + c = -6

-400 + c = -6

c = -6 + 400

c = 394

You can graph y = 16x² + 160x + 394 at Desmos.com to confirm the given vertex.

I hope this helps

Answer 3

y = 16(x² + b/16x + b²/32²) + c - b²/64 = 0; y = 16(x + b/32)² + C - b²/64; - b/32 = - 5; b = 160

c - b²/64 = - 6; c = 160²/64 - 6 = 394

So, y = 16x² + 160x + 394