SOLUTION: Find b and c so that y=-15x^2+bx+c has vertex (-10,0)?

Question 1135550: Find b and c so that y=-15x^2+bx+c has vertex (-10,0)?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

The given equation is in the form
y = ax^2 + bx + c
We see that only a = -15 is known while b and c are currently unknown

Vertex = (h,k) = (-10,0)
h = -10 and k = 0

Vertex Form:
y = a(x-h)^2 + k
y = -15(x-(-10))^2 + 0 ... plug in the given values
y = -15(x+10)^2
y = -15(x+10)(x+10)
y = -15(x^2+10x+10x+100) ... FOIL rule
y = -15(x^2+20x+100)
y = -15(x^2)-15(20x)-15(100) ... distribute
y = -15x^2-300x-1500

Compare that last equation with the form y = ax^2+bx+c and we see that
a = -15
b = -300
c = -1500

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Answers:
b = -300
c = -1500

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