Find the Maximum/Minimum Value f(x)=-x^2-11x-26

$f (x) = - x^{2} - 11 x - 26$

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 $b$ .

$x = - \frac{- 11}{2 (- 1)}$

Remove parentheses.

$x = - \frac{- 11}{2 (- 1)}$

Simplify $- \frac{- 11}{2 (- 1)}$ .

Multiply $2$ by $- 1$ .

$x = - \frac{- 11}{- 2}$

Dividing two negative values results in a positive value.

$x = - \frac{11}{2}$

Step 3

Evaluate $f (- \frac{11}{2})$ .

Replace the variable $x$ with $- \frac{11}{2}$ in the expression.

$f (- \frac{11}{2}) = - {(- \frac{11}{2})}^{2} - 11 (- \frac{11}{2}) - 26$

Simplify the result.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{11}{2}$ .

$f (- \frac{11}{2}) = - ({(- 1)}^{2} {(\frac{11}{2})}^{2}) - 11 (- \frac{11}{2}) - 26$

Apply the product rule to $\frac{11}{2}$ .

$f (- \frac{11}{2}) = - ({(- 1)}^{2} (\frac{11^{2}}{2^{2}})) - 11 (- \frac{11}{2}) - 26$

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

Move ${(- 1)}^{2}$ .

$f (- \frac{11}{2}) = {(- 1)}^{2} \cdot (- 1 \frac{11^{2}}{2^{2}}) - 11 (- \frac{11}{2}) - 26$

Multiply ${(- 1)}^{2}$ by $- 1$ .

Raise $- 1$ to the power of $1$ .

$f (- \frac{11}{2}) = {(- 1)}^{2} \cdot ((- 1) (\frac{11^{2}}{2^{2}})) - 11 (- \frac{11}{2}) - 26$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (- \frac{11}{2}) = {(- 1)}^{2 + 1} (\frac{11^{2}}{2^{2}}) - 11 (- \frac{11}{2}) - 26$

Add $2$ and $1$ .

$f (- \frac{11}{2}) = {(- 1)}^{3} (\frac{11^{2}}{2^{2}}) - 11 (- \frac{11}{2}) - 26$

Raise $- 1$ to the power of $3$ .

$f (- \frac{11}{2}) = - \frac{11^{2}}{2^{2}} - 11 (- \frac{11}{2}) - 26$

Raise $11$ to the power of $2$ .

$f (- \frac{11}{2}) = - \frac{121}{2^{2}} - 11 (- \frac{11}{2}) - 26$

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

$f (- \frac{11}{2}) = - \frac{121}{4} - 11 (- \frac{11}{2}) - 26$

Multiply $- 11 (- \frac{11}{2})$ .

Multiply $- 1$ by $- 11$ .

$f (- \frac{11}{2}) = - \frac{121}{4} + 11 (\frac{11}{2}) - 26$

Combine $11$ and $\frac{11}{2}$ .

$f (- \frac{11}{2}) = - \frac{121}{4} + \frac{11 \cdot 11}{2} - 26$

Multiply $11$ by $11$ .

$f (- \frac{11}{2}) = - \frac{121}{4} + \frac{121}{2} - 26$

Find the common denominator.

Multiply $\frac{121}{2}$ by $\frac{2}{2}$ .

$f (- \frac{11}{2}) = - \frac{121}{4} + \frac{121}{2} \cdot \frac{2}{2} - 26$

Multiply $\frac{121}{2}$ by $\frac{2}{2}$ .

$f (- \frac{11}{2}) = - \frac{121}{4} + \frac{121 \cdot 2}{2 \cdot 2} - 26$

Write $- 26$ as a fraction with denominator $1$ .

$f (- \frac{11}{2}) = - \frac{121}{4} + \frac{121 \cdot 2}{2 \cdot 2} + \frac{- 26}{1}$

Multiply $\frac{- 26}{1}$ by $\frac{4}{4}$ .

$f (- \frac{11}{2}) = - \frac{121}{4} + \frac{121 \cdot 2}{2 \cdot 2} + \frac{- 26}{1} \cdot \frac{4}{4}$

Multiply $\frac{- 26}{1}$ by $\frac{4}{4}$ .

$f (- \frac{11}{2}) = - \frac{121}{4} + \frac{121 \cdot 2}{2 \cdot 2} + \frac{- 26 \cdot 4}{4}$

Multiply $2$ by $2$ .

$f (- \frac{11}{2}) = - \frac{121}{4} + \frac{121 \cdot 2}{4} + \frac{- 26 \cdot 4}{4}$

Combine the numerators over the common denominator.

$f (- \frac{11}{2}) = \frac{- 121 + 121 \cdot 2 - 26 \cdot 4}{4}$

Simplify each term.

Multiply $121$ by $2$ .

$f (- \frac{11}{2}) = \frac{- 121 + 242 - 26 \cdot 4}{4}$

Multiply $- 26$ by $4$ .

$f (- \frac{11}{2}) = \frac{- 121 + 242 - 104}{4}$

Simplify by adding and subtracting.

Add $- 121$ and $242$ .

$f (- \frac{11}{2}) = \frac{121 - 104}{4}$

Subtract $104$ from $121$ .

$f (- \frac{11}{2}) = \frac{17}{4}$

The final answer is $\frac{17}{4}$ .

$\frac{17}{4}$

Step 4

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

$(- \frac{11}{2}, \frac{17}{4})$

Step 5

Find the Maximum/Minimum Value f(x)=-x^2-11x-26

Related Questions

Support LCMGCF.com by disabling your adblocker.