Find the Maximum/Minimum Value 2x^2+10x+9

$2 x^{2} + 10 x + 9$

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 $b$ .

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

Remove parentheses.

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

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

Factor $2$ out of $10$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 3

Evaluate $f (- \frac{5}{2})$ .

Replace the variable $x$ with $- \frac{5}{2}$ in the expression.

$f (- \frac{5}{2}) = 2 {(- \frac{5}{2})}^{2} + 10 (- \frac{5}{2}) + 9$

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 $- \frac{5}{2}$ .

$f (- \frac{5}{2}) = 2 ({(- 1)}^{2} {(\frac{5}{2})}^{2}) + 10 (- \frac{5}{2}) + 9$

Apply the product rule to $\frac{5}{2}$ .

$f (- \frac{5}{2}) = 2 ({(- 1)}^{2} (\frac{5^{2}}{2^{2}})) + 10 (- \frac{5}{2}) + 9$

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

$f (- \frac{5}{2}) = 2 (1 (\frac{5^{2}}{2^{2}})) + 10 (- \frac{5}{2}) + 9$

Multiply $\frac{5^{2}}{2^{2}}$ by $1$ .

$f (- \frac{5}{2}) = 2 (\frac{5^{2}}{2^{2}}) + 10 (- \frac{5}{2}) + 9$

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

$f (- \frac{5}{2}) = 2 (\frac{25}{2^{2}}) + 10 (- \frac{5}{2}) + 9$

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

$f (- \frac{5}{2}) = 2 (\frac{25}{4}) + 10 (- \frac{5}{2}) + 9$

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$f (- \frac{5}{2}) = 2 (\frac{25}{2 (2)}) + 10 (- \frac{5}{2}) + 9$

Cancel the common factor.

$f (- \frac{5}{2}) = 2 (\frac{25}{2 \cdot 2}) + 10 (- \frac{5}{2}) + 9$

Rewrite the expression.

$f (- \frac{5}{2}) = \frac{25}{2} + 10 (- \frac{5}{2}) + 9$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{5}{2}$ into the numerator.

$f (- \frac{5}{2}) = \frac{25}{2} + 10 (\frac{- 5}{2}) + 9$

Factor $2$ out of $10$ .

$f (- \frac{5}{2}) = \frac{25}{2} + 2 (5) (\frac{- 5}{2}) + 9$

Cancel the common factor.

$f (- \frac{5}{2}) = \frac{25}{2} + 2 \cdot (5 (\frac{- 5}{2})) + 9$

Rewrite the expression.

$f (- \frac{5}{2}) = \frac{25}{2} + 5 \cdot - 5 + 9$

Multiply $5$ by $- 5$ .

$f (- \frac{5}{2}) = \frac{25}{2} - 25 + 9$

Find the common denominator.

Write $- 25$ as a fraction with denominator $1$ .

$f (- \frac{5}{2}) = \frac{25}{2} + \frac{- 25}{1} + 9$

Multiply $\frac{- 25}{1}$ by $\frac{2}{2}$ .

$f (- \frac{5}{2}) = \frac{25}{2} + \frac{- 25}{1} \cdot \frac{2}{2} + 9$

Multiply $\frac{- 25}{1}$ by $\frac{2}{2}$ .

$f (- \frac{5}{2}) = \frac{25}{2} + \frac{- 25 \cdot 2}{2} + 9$

Write $9$ as a fraction with denominator $1$ .

$f (- \frac{5}{2}) = \frac{25}{2} + \frac{- 25 \cdot 2}{2} + \frac{9}{1}$

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

$f (- \frac{5}{2}) = \frac{25}{2} + \frac{- 25 \cdot 2}{2} + \frac{9}{1} \cdot \frac{2}{2}$

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

$f (- \frac{5}{2}) = \frac{25}{2} + \frac{- 25 \cdot 2}{2} + \frac{9 \cdot 2}{2}$

Combine the numerators over the common denominator.

$f (- \frac{5}{2}) = \frac{25 - 25 \cdot 2 + 9 \cdot 2}{2}$

Simplify each term.

Multiply $- 25$ by $2$ .

$f (- \frac{5}{2}) = \frac{25 - 50 + 9 \cdot 2}{2}$

Multiply $9$ by $2$ .

$f (- \frac{5}{2}) = \frac{25 - 50 + 18}{2}$

Simplify the expression.

Subtract $50$ from $25$ .

$f (- \frac{5}{2}) = \frac{- 25 + 18}{2}$

Add $- 25$ and $18$ .

$f (- \frac{5}{2}) = \frac{- 7}{2}$

Move the negative in front of the fraction.

$f (- \frac{5}{2}) = - \frac{7}{2}$

The final answer is $- \frac{7}{2}$ .

$- \frac{7}{2}$

Step 4

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

$(- \frac{5}{2}, - \frac{7}{2})$

Step 5

Find the Maximum/Minimum Value 2x^2+10x+9

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