Question

How do you find the max or minimum of `f(x)=3/4x^2-5x-2`?

Answer 1

We have a minimum of `-10 1/3` at `x=10/3`.

Explanation:

To find

first that whether `f(x)=3/4x^2-5x-2` has a maximum or minimum, one has to first look at the coefficient of `x^2`. Here it is positive, so we have a minimum . Note that for this function has to be in the form `f(x)=ax^2+bx+c`.

Secondly to find the minimum value (or maximum in case coefficient of `x^2` is negative), we should convert it into vertex form i.e. `f(x)=a(x-h)^2+k`, where we have a maximum of `k` at `x=h`.

`f(x)=3/4x^2-5x-2`

= `3/4(x^2-20/3x)-2`

= `3/4(x^2-2xx10/3x+(10/3)^2)-(10/3)^2xx3/4-2`

= `3/4(x-10/3)^2-100/12-2`

= `3/4(x-10/3)^2-31/3`

Hence, we have a minimum of `-31/3=-10 1/3` at `x=10/3`.

graph{3/4x^2-5x-2 [-17.41, 22.59, -11.92, 8.08]}