Find the Maximum/Minimum Value f(x)=4x^2-24x+41

$f (x) = 4 x^{2} - 24 x + 41$

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 $b$ .

$x = - \frac{- 24}{2 (4)}$

Remove parentheses.

$x = - \frac{- 24}{2 (4)}$

Simplify $- \frac{- 24}{2 (4)}$ .

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

Factor $2$ out of $- 24$ .

$x = - \frac{2 \cdot - 12}{2 \cdot 4}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 4$ .

$x = - \frac{2 \cdot - 12}{2 (4)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 12}{2 \cdot 4}$

Rewrite the expression.

$x = - \frac{- 12}{4}$

Cancel the common factor of $- 12$ and $4$ .

Factor $4$ out of $- 12$ .

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

Cancel the common factors.

Factor $4$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 3$ by $1$ .

$x = - - 3$

Multiply $- 1$ by $- 3$ .

$x = 3$

Step 3

Evaluate $f (3)$ .

Replace the variable $x$ with $3$ in the expression.

$f (3) = 4 {(3)}^{2} - 24 \cdot 3 + 41$

Simplify the result.

Simplify each term.

Raise $3$ to the power of $2$ .

$f (3) = 4 \cdot 9 - 24 \cdot 3 + 41$

Multiply $4$ by $9$ .

$f (3) = 36 - 24 \cdot 3 + 41$

Multiply $- 24$ by $3$ .

$f (3) = 36 - 72 + 41$

Simplify by adding and subtracting.

Subtract $72$ from $36$ .

$f (3) = - 36 + 41$

Add $- 36$ and $41$ .

$f (3) = 5$

The final answer is $5$ .

$5$

Step 4

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

$(3, 5)$

Step 5

Find the Maximum/Minimum Value f(x)=4x^2-24x+41

Related Questions

Support LCMGCF.com by disabling your adblocker.