Find the Maximum/Minimum Value f(x)=3x^2-24x+53

$f (x) = 3 x^{2} - 24 x + 53$

Step 1

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

$f_{min}$ $x = a x^{2} + b x + c$ 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 $- 12$ .

$x = - \frac{3 \cdot - 4}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

$x = - \frac{3 \cdot - 4}{3 (1)}$

Cancel the common factor.

$x = - \frac{3 \cdot - 4}{3 \cdot 1}$

Rewrite the expression.

$x = - \frac{- 4}{1}$

Divide $- 4$ by $1$ .

$x = - - 4$

Multiply $- 1$ by $- 4$ .

$x = 4$

Step 3

Evaluate $f (4)$ .

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

$f (4) = 3 {(4)}^{2} - 24 \cdot 4 + 53$

Simplify the result.

Simplify each term.

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

$f (4) = 3 \cdot 16 - 24 \cdot 4 + 53$

Multiply $3$ by $16$ .

$f (4) = 48 - 24 \cdot 4 + 53$

Multiply $- 24$ by $4$ .

$f (4) = 48 - 96 + 53$

Simplify by adding and subtracting.

Subtract $96$ from $48$ .

$f (4) = - 48 + 53$

Add $- 48$ and $53$ .

$f (4) = 5$

The final answer is $5$ .

$5$

Step 4

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

$(4, 5)$

Step 5

Find the Maximum/Minimum Value f(x)=3x^2-24x+53

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