Find the Maximum/Minimum Value f(x)=5x^2+50x+121

$f (x) = 5 x^{2} + 50 x + 121$

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 $b$ .

$x = - \frac{50}{2 (5)}$

Remove parentheses.

$x = - \frac{50}{2 (5)}$

Simplify $- \frac{50}{2 (5)}$ .

Cancel the common factor of $50$ and $2$ .

Factor $2$ out of $50$ .

$x = - \frac{2 \cdot 25}{2 \cdot 5}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 5$ .

$x = - \frac{2 \cdot 25}{2 (5)}$

Cancel the common factor.

$x = - \frac{2 \cdot 25}{2 \cdot 5}$

Rewrite the expression.

$x = - \frac{25}{5}$

Cancel the common factor of $25$ and $5$ .

Factor $5$ out of $25$ .

$x = - \frac{5 \cdot 5}{5}$

Cancel the common factors.

Factor $5$ out of $5$ .

$x = - \frac{5 \cdot 5}{5 (1)}$

Cancel the common factor.

$x = - \frac{5 \cdot 5}{5 \cdot 1}$

Rewrite the expression.

$x = - \frac{5}{1}$

Divide $5$ by $1$ .

$x = - 1 \cdot 5$

Multiply $- 1$ by $5$ .

$x = - 5$

Step 3

Evaluate $f (- 5)$ .

Replace the variable $x$ with $- 5$ in the expression.

$f (- 5) = 5 {(- 5)}^{2} + 50 (- 5) + 121$

Simplify the result.

Simplify each term.

Raise $- 5$ to the power of $2$ .

$f (- 5) = 5 \cdot 25 + 50 (- 5) + 121$

Multiply $5$ by $25$ .

$f (- 5) = 125 + 50 (- 5) + 121$

Multiply $50$ by $- 5$ .

$f (- 5) = 125 - 250 + 121$

Simplify by adding and subtracting.

Subtract $250$ from $125$ .

$f (- 5) = - 125 + 121$

Add $- 125$ and $121$ .

$f (- 5) = - 4$

The final answer is $- 4$ .

$- 4$

Step 4

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

$(- 5, - 4)$

Step 5

Find the Maximum/Minimum Value f(x)=5x^2+50x+121

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