Find the Maximum/Minimum Value y=1/3x^3-5/2x^2+6x-8

$y = \frac{1}{3} x^{3} - \frac{5}{2} x^{2} + 6 x - 8$

Simplify each term.

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Combine $\frac{1}{3}$ and $x^{3}$ .

$y = \frac{x^{3}}{3} - \frac{5}{2} x^{2} + 6 x - 8$

Combine $x^{2}$ and $\frac{5}{2}$ .

$y = \frac{x^{3}}{3} - \frac{x^{2} \cdot 5}{2} + 6 x - 8$

Move $5$ to the left of $x^{2}$ .

$y = \frac{x^{3}}{3} - \frac{5 x^{2}}{2} + 6 x - 8$

The minimum and maximum cannot be determined because this is not a quadratic function.

Not a Quadratic Function

Find the Maximum/Minimum Value y=1/3x^3-5/2x^2+6x-8

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