Find the Maximum/Minimum Value f(x)=2x^2+20x+53

$f (x) = 2 x^{2} + 20 x + 53$

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 $b$ .

$x = - \frac{20}{2 (2)}$

Remove parentheses.

$x = - \frac{20}{2 (2)}$

Simplify $- \frac{20}{2 (2)}$ .

Cancel the common factor of $20$ and $2$ .

Factor $2$ out of $20$ .

$x = - \frac{2 \cdot 10}{2 \cdot 2}$

Cancel the common factors.

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

$x = - \frac{2 \cdot 10}{2 (2)}$

Cancel the common factor.

$x = - \frac{2 \cdot 10}{2 \cdot 2}$

Rewrite the expression.

$x = - \frac{10}{2}$

Cancel the common factor of $10$ and $2$ .

Factor $2$ out of $10$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

$x = - \frac{2 \cdot 5}{2 (1)}$

Cancel the common factor.

$x = - \frac{2 \cdot 5}{2 \cdot 1}$

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.

$f (- 5) = 2 {(- 5)}^{2} + 20 (- 5) + 53$

Simplify the result.

Simplify each term.

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

$f (- 5) = 2 \cdot 25 + 20 (- 5) + 53$

Multiply $2$ by $25$ .

$f (- 5) = 50 + 20 (- 5) + 53$

Multiply $20$ by $- 5$ .

$f (- 5) = 50 - 100 + 53$

Simplify by adding and subtracting.

Subtract $100$ from $50$ .

$f (- 5) = - 50 + 53$

Add $- 50$ and $53$ .

$f (- 5) = 3$

The final answer is $3$ .

$3$

Step 4

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

$(- 5, 3)$

Step 5

Find the Maximum/Minimum Value f(x)=2x^2+20x+53

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