# What is the maximum value of y = 3x - 18x^2 + 1

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Find the maximum value for the function `y=3x-18x^2+1 `

Without calculus:

Put in standard form and recognize the graph of the function is a parabola opening down. The maximum occurs at the vertex. The vertex lies on the axis of symmetry: for `y=ax^2+bx+c ` the axis of symmetry is `x=(-b)/(2a) ` .

` `In standard form we have `y=-18x^2+3x+1 ` so the axis of symmetry is `x=(-3)/(2(-18))=1/12 ` .

The value of y when `x=1/12 ` is `y=-18(1/12)^2+3(1/12)+1 `

`y=(-18)/144+3/12+1=-1/8+1/4+1=9/8 `

The extreme values of a function y = f(x) lies at points where f'(x) = 0. If a is a solution of f'(x) = 0 and f''(a) is negative, the function has a maxima at the point.

For the function y = 3x - 18x^2 + 1, y' = 3 - 36x

y' = 0

=> 3 - 36x = 0

=> x = 1/12

y'' = -36 which is negative for x = 1/12

At x = 1/12, y = 3*(1/12) - 18*(1/12)^2 + 1 = 9/8

The maximum value of the function is 9/8.