Find the Maximum/Minimum Value f(x)=x^2+9x+18

$f (x) = x^{2} + 9 x + 18$

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 $b$ .

$x = - \frac{9}{2 (1)}$

Remove parentheses.

$x = - \frac{9}{2 (1)}$

Multiply $2$ by $1$ .

$x = - \frac{9}{2}$

Step 3

Evaluate $f (- \frac{9}{2})$ .

Replace the variable $x$ with $- \frac{9}{2}$ in the expression.

$f (- \frac{9}{2}) = {(- \frac{9}{2})}^{2} + 9 (- \frac{9}{2}) + 18$

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 $- \frac{9}{2}$ .

$f (- \frac{9}{2}) = {(- 1)}^{2} {(\frac{9}{2})}^{2} + 9 (- \frac{9}{2}) + 18$

Apply the product rule to $\frac{9}{2}$ .

$f (- \frac{9}{2}) = {(- 1)}^{2} (\frac{9^{2}}{2^{2}}) + 9 (- \frac{9}{2}) + 18$

Raise $- 1$ to the power of $2$ .

$f (- \frac{9}{2}) = 1 (\frac{9^{2}}{2^{2}}) + 9 (- \frac{9}{2}) + 18$

Multiply $\frac{9^{2}}{2^{2}}$ by $1$ .

$f (- \frac{9}{2}) = \frac{9^{2}}{2^{2}} + 9 (- \frac{9}{2}) + 18$

Raise $9$ to the power of $2$ .

$f (- \frac{9}{2}) = \frac{81}{2^{2}} + 9 (- \frac{9}{2}) + 18$

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

$f (- \frac{9}{2}) = \frac{81}{4} + 9 (- \frac{9}{2}) + 18$

Multiply $9 (- \frac{9}{2})$ .

Multiply $- 1$ by $9$ .

$f (- \frac{9}{2}) = \frac{81}{4} - 9 (\frac{9}{2}) + 18$

Combine $- 9$ and $\frac{9}{2}$ .

$f (- \frac{9}{2}) = \frac{81}{4} + \frac{- 9 \cdot 9}{2} + 18$

Multiply $- 9$ by $9$ .

$f (- \frac{9}{2}) = \frac{81}{4} + \frac{- 81}{2} + 18$

Move the negative in front of the fraction.

$f (- \frac{9}{2}) = \frac{81}{4} - \frac{81}{2} + 18$

Find the common denominator.

Multiply $\frac{81}{2}$ by $\frac{2}{2}$ .

$f (- \frac{9}{2}) = \frac{81}{4} - (\frac{81}{2} \cdot \frac{2}{2}) + 18$

Multiply $\frac{81}{2}$ by $\frac{2}{2}$ .

$f (- \frac{9}{2}) = \frac{81}{4} - \frac{81 \cdot 2}{2 \cdot 2} + 18$

Write $18$ as a fraction with denominator $1$ .

$f (- \frac{9}{2}) = \frac{81}{4} - \frac{81 \cdot 2}{2 \cdot 2} + \frac{18}{1}$

Multiply $\frac{18}{1}$ by $\frac{4}{4}$ .

$f (- \frac{9}{2}) = \frac{81}{4} - \frac{81 \cdot 2}{2 \cdot 2} + \frac{18}{1} \cdot \frac{4}{4}$

Multiply $\frac{18}{1}$ by $\frac{4}{4}$ .

$f (- \frac{9}{2}) = \frac{81}{4} - \frac{81 \cdot 2}{2 \cdot 2} + \frac{18 \cdot 4}{4}$

Multiply $2$ by $2$ .

$f (- \frac{9}{2}) = \frac{81}{4} - \frac{81 \cdot 2}{4} + \frac{18 \cdot 4}{4}$

Combine the numerators over the common denominator.

$f (- \frac{9}{2}) = \frac{81 - 81 \cdot 2 + 18 \cdot 4}{4}$

Simplify each term.

Multiply $- 81$ by $2$ .

$f (- \frac{9}{2}) = \frac{81 - 162 + 18 \cdot 4}{4}$

Multiply $18$ by $4$ .

$f (- \frac{9}{2}) = \frac{81 - 162 + 72}{4}$

Simplify the expression.

Subtract $162$ from $81$ .

$f (- \frac{9}{2}) = \frac{- 81 + 72}{4}$

Add $- 81$ and $72$ .

$f (- \frac{9}{2}) = \frac{- 9}{4}$

Move the negative in front of the fraction.

$f (- \frac{9}{2}) = - \frac{9}{4}$

The final answer is $- \frac{9}{4}$ .

$- \frac{9}{4}$

Step 4

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

$(- \frac{9}{2}, - \frac{9}{4})$

Step 5

Find the Maximum/Minimum Value f(x)=x^2+9x+18

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