Find the Maximum/Minimum Value -3x^2+24x-51

$- 3 x^{2} + 24 x - 51$

Step 1

The maximum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is negative, the maximum value of the function is $f (- \frac{b}{2 a})$ .

$f_{max}$ $x = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$

Step 2

Find the value of $x = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

$x = - \frac{24}{2 (- 3)}$

Remove parentheses.

$x = - \frac{24}{2 (- 3)}$

Simplify $- \frac{24}{2 (- 3)}$ .

Cancel the common factor of $24$ and $2$ .

Factor $2$ out of $24$ .

$x = - \frac{2 \cdot 12}{2 \cdot - 3}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 3$ .

$x = - \frac{2 \cdot 12}{2 (- 3)}$

Cancel the common factor.

$x = - \frac{2 \cdot 12}{2 \cdot - 3}$

Rewrite the expression.

$x = - \frac{12}{- 3}$

Cancel the common factor of $12$ and $- 3$ .

Factor $3$ out of $12$ .

$x = - \frac{3 (4)}{- 3}$

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$x = - (- 1 \cdot 4)$

Multiply $- (- 1 \cdot 4)$ .

Multiply $- 1$ by $4$ .

$x = - - 4$

Multiply $- 1$ by $- 4$ .

$x = 4$

Step 3

Evaluate $f (4)$ .

Replace the variable $x$ with $4$ in the expression.

$f (4) = - 3 {(4)}^{2} + 24 (4) - 51$

Simplify the result.

Simplify each term.

Raise $4$ to the power of $2$ .

$f (4) = - 3 \cdot 16 + 24 (4) - 51$

Multiply $- 3$ by $16$ .

$f (4) = - 48 + 24 (4) - 51$

Multiply $24$ by $4$ .

$f (4) = - 48 + 96 - 51$

Simplify by adding and subtracting.

Add $- 48$ and $96$ .

$f (4) = 48 - 51$

Subtract $51$ from $48$ .

$f (4) = - 3$

The final answer is $- 3$ .

$- 3$

Step 4

Use the $x$ and $y$ values to find where the maximum occurs.

$(4, - 3)$

Step 5

Find the Maximum/Minimum Value -3x^2+24x-51

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