# What is m if the minim of function y = 3x^2-mx+11 is -4?

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You may use different approaches to evaluate `m` , either using derivatives, or using the vertex formula `(-b/(2a),(4ac-b^2)/(4a))` .

Using derivative approach, you need to remember that the function reaches its maximum or minimum at `x = x_0` , where `x_0` represents the solution to equation `f'(x) = 0.`

Evaluating the derivative of the given function, yields:

`y' = 6x - m`

You need to set up the equation ` y' = 0` , such that:

`6x - m = 0 => 6x = m => x = m/6`

Hence, the function reaches its minimum `y = -4 ` at `x = m/6` , thus, you need to evaluate the function at `x = m/6` , such that:

`-4 = 3*m^2/36 - m*m/6 + 11`

`-4*36 = 3m^2 - 6m^2 + 36*11 => -4*36 - 36*11 = -3m^2`

`36(4 + 11) = 3m^2 => 12*15 = m^2 => m = +-6sqrt(5)`

Hence, evaluating the values of m, using derivatives, yields `m = +-6sqrt(5).`

You may also use the vertex approach, since the problem provides that the y value of vertex (minimum) is of `-4` , such that:

`-4 = (4ac-b^2)/(4a)`

Identifying the coefficients `a,b,c` yields:

`-16*3 = 12*11 - m^2 => m^2 = 12*11 + 16*3`

`m^2 = 132 + 48 = 180 => m = +-6sqrt5`

**Hence, using both approaches, yields **`m = +-6sqrt5.`