Find the Maximum/Minimum Value y=-x^2-x+6

$y = - x^{2} - x + 6$

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 $b$ .

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

Remove parentheses.

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

Dividing two negative values results in a positive value.

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

Step 3

Evaluate $f (- \frac{1}{2})$ .

Replace the variable $x$ with $- \frac{1}{2}$ in the expression.

$f (- \frac{1}{2}) = - {(- \frac{1}{2})}^{2} - (- \frac{1}{2}) + 6$

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 $- \frac{1}{2}$ .

$f (- \frac{1}{2}) = - ({(- 1)}^{2} {(\frac{1}{2})}^{2}) - (- \frac{1}{2}) + 6$

Apply the product rule to $\frac{1}{2}$ .

$f (- \frac{1}{2}) = - ({(- 1)}^{2} (\frac{1^{2}}{2^{2}})) - (- \frac{1}{2}) + 6$

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

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

$f (- \frac{1}{2}) = {(- 1)}^{2} \cdot (- 1 \frac{1^{2}}{2^{2}}) - (- \frac{1}{2}) + 6$

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

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

$f (- \frac{1}{2}) = {(- 1)}^{2} \cdot ((- 1) (\frac{1^{2}}{2^{2}})) - (- \frac{1}{2}) + 6$

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

$f (- \frac{1}{2}) = {(- 1)}^{2 + 1} (\frac{1^{2}}{2^{2}}) - (- \frac{1}{2}) + 6$

Add $2$ and $1$ .

$f (- \frac{1}{2}) = {(- 1)}^{3} (\frac{1^{2}}{2^{2}}) - (- \frac{1}{2}) + 6$

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

$f (- \frac{1}{2}) = - \frac{1^{2}}{2^{2}} - (- \frac{1}{2}) + 6$

One to any power is one.

$f (- \frac{1}{2}) = - \frac{1}{2^{2}} - (- \frac{1}{2}) + 6$

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

$f (- \frac{1}{2}) = - \frac{1}{4} - (- \frac{1}{2}) + 6$

Multiply $- (- \frac{1}{2})$ .

Multiply $- 1$ by $- 1$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + 1 (\frac{1}{2}) + 6$

Multiply $\frac{1}{2}$ by $1$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{1}{2} + 6$

Find the common denominator.

Multiply $\frac{1}{2}$ by $\frac{2}{2}$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{1}{2} \cdot \frac{2}{2} + 6$

Multiply $\frac{1}{2}$ by $\frac{2}{2}$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{2}{2 \cdot 2} + 6$

Write $6$ as a fraction with denominator $1$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{2}{2 \cdot 2} + \frac{6}{1}$

Multiply $\frac{6}{1}$ by $\frac{4}{4}$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{2}{2 \cdot 2} + \frac{6}{1} \cdot \frac{4}{4}$

Multiply $\frac{6}{1}$ by $\frac{4}{4}$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{2}{2 \cdot 2} + \frac{6 \cdot 4}{4}$

Multiply $2$ by $2$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{2}{4} + \frac{6 \cdot 4}{4}$

Combine the numerators over the common denominator.

$f (- \frac{1}{2}) = \frac{- 1 + 2 + 6 \cdot 4}{4}$

Simplify the expression.

Multiply $6$ by $4$ .

$f (- \frac{1}{2}) = \frac{- 1 + 2 + 24}{4}$

Add $- 1$ and $2$ .

$f (- \frac{1}{2}) = \frac{1 + 24}{4}$

Add $1$ and $24$ .

$f (- \frac{1}{2}) = \frac{25}{4}$

The final answer is $\frac{25}{4}$ .

$\frac{25}{4}$

Step 4

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

$(- \frac{1}{2}, \frac{25}{4})$

Step 5

Find the Maximum/Minimum Value y=-x^2-x+6

Related Questions

Support LCMGCF.com by disabling your adblocker.