SOLUTION: I am in over my head here, so please use very simple explanations. Find b and c so that the parabola y=20x^2+bx+c has vertex (10,-3)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: I am in over my head here, so please use very simple explanations. Find b and c so that the parabola y=20x^2+bx+c has vertex (10,-3)      Log On


   

Question 178427: I am in over my head here, so please use very simple explanations. Find b and c so that the parabola y=20x^2+bx+c has vertex (10,-3)
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
the vertex is on the axis of symmetry

the general equation for the axis of symmetry is x = -b/(2a)

in this case, 10=-b/(2*20)

multiplying by 40 __ 400=-b __ -400=b

the value of c can be found by substituting the vertex coordinates

-3=20(10^2)-400(10)+c __ -3=2000-4000+c

adding 2000 __ 1997=c