SOLUTION: find the vertex of y=(-1/4)x^2-(1/5)x

Question 632691: find the vertex of y=(-1/4)x^2-(1/5)x
Answer by dfrazzetto(283) (Show Source): You can put this solution on YOUR website!
coefficient of the x^2 term is negative, so it's an upside down parabola, vertex will be at the local maximum, which occurs when the derivative of y(x) = 0
y(x) = -x^2/4 - x/5
dy/dx (y(x)) = -x/2 - 1/5
set equal to zero:
-x/2 - 1/5 = 0
-x/2 = 1/5
x = -2/5
The highest point on the upside down parabola, which is the vertex, occurs when x= -2/5
y(x) = -x^2/4 - x/5
y(-2/5) = (-1/4)(-2/5)^2 - (-2/5)/5
= (-1/4)(4/25) + 2/25
= -1/25 + 2/25 = 1/25
x = -2/5, y = 1/25
(-2/5, 1/25)
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