Question

What is the minimum or maximum of `j(x)=-5/3x^2+9/10x+21/4`?

Answer 1

Since the coefficient of `x^2` is negative we are talking about a maximum.

Explanation:

To find it, we set the derivative of `j(x)` equal to `0`

`j'(x)=-2*5/3x+9/10=0`

`->9/10=10/3x->x=9/10-:10/3=9/10*3/10=27/100`

So this is the `x`-value of the maximum. If you put that in the original formula, you will find the corresponding `j(x)`-value.
graph{-5/3x^2+.9x+21/4 [-7.24, 8.564, -1.073, 6.83]}