Find b and c so that y=5x^2+bx+c and has a vertex of (8.-10)

Question

Its precalculus. And I want toknow what the value of B and C would be for the equation y=5x^2+bx+c with a vertex of (8.-10).

Answer 1

y = 5x^2+bx+c ---> (1)

Parabola equation y = ax^2+bx+c

Vertex of parabola (x,y) = (8,-10)

a = 5

Axis of symmetry x = -b/2a = -b/10

-b/10 = 8

Multiple to each side by 10.

-b = 80

Multiple to each side by negitive one.

b = -80

Substitute b,x value in (1).

5x^2-80x+c = -10

5(8)^2-80(8)+c = -10

320-640+c = -10

c-320 = -10

Add 320 to each side.

c = 310

b = -80, c = 310.

 

Comments

very good, correct answer.