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

Aug 6, 2015

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

#### Explanation:

To find it, we set the derivative of $j \left(x\right)$ equal to $0$

$j ' \left(x\right) = - 2 \cdot \frac{5}{3} x + \frac{9}{10} = 0$

$\to \frac{9}{10} = \frac{10}{3} x \to x = \frac{9}{10} \div \frac{10}{3} = \frac{9}{10} \cdot \frac{3}{10} = \frac{27}{100}$

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