Find the Maximum/Minimum Value -3-x+x^2

$- 3 - x + x^{2}$

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 $x = - \frac{b}{2 a}$

Find the value of $x = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

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

Remove parentheses.

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

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

Cancel the common factor of $- 1$ and $1$ .

Rewrite $- 1$ as $- 1 (1)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

Multiply $- 1$ by $- 1$ .

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

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

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

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

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

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

Simplify the result.

Simplify each term.

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

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

One to any power is one.

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

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

$f (\frac{1}{2}) = - 3 - \frac{1}{2} + \frac{1}{4}$

Find the common denominator.

Write $- 3$ as a fraction with denominator $1$ .

$f (\frac{1}{2}) = \frac{- 3}{1} - \frac{1}{2} + \frac{1}{4}$

Multiply $\frac{- 3}{1}$ by $\frac{4}{4}$ .

$f (\frac{1}{2}) = \frac{- 3}{1} \cdot \frac{4}{4} - \frac{1}{2} + \frac{1}{4}$

Multiply $\frac{- 3}{1}$ by $\frac{4}{4}$ .

$f (\frac{1}{2}) = \frac{- 3 \cdot 4}{4} - \frac{1}{2} + \frac{1}{4}$

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

$f (\frac{1}{2}) = \frac{- 3 \cdot 4}{4} - (\frac{1}{2} \cdot \frac{2}{2}) + \frac{1}{4}$

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

$f (\frac{1}{2}) = \frac{- 3 \cdot 4}{4} - \frac{2}{2 \cdot 2} + \frac{1}{4}$

Multiply $2$ by $2$ .

$f (\frac{1}{2}) = \frac{- 3 \cdot 4}{4} - \frac{2}{4} + \frac{1}{4}$

Combine the numerators over the common denominator.

$f (\frac{1}{2}) = \frac{- 3 \cdot 4 - 2 + 1}{4}$

Simplify the expression.

Multiply $- 3$ by $4$ .

$f (\frac{1}{2}) = \frac{- 12 - 2 + 1}{4}$

Subtract $2$ from $- 12$ .

$f (\frac{1}{2}) = \frac{- 14 + 1}{4}$

Add $- 14$ and $1$ .

$f (\frac{1}{2}) = \frac{- 13}{4}$

Move the negative in front of the fraction.

$f (\frac{1}{2}) = - \frac{13}{4}$

The final answer is $- \frac{13}{4}$ .

$- \frac{13}{4}$

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

$(\frac{1}{2}, - \frac{13}{4})$

Find the Maximum/Minimum Value -3-x+x^2

Related Questions

Support LCMGCF.com by disabling your adblocker.