Find the Maximum/Minimum Value 2x^2+6x+3

$2 x^{2} + 6 x + 3$

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 $b$ .

$x = - \frac{6}{2 (2)}$

Remove parentheses.

$x = - \frac{6}{2 (2)}$

Cancel the common factor of $6$ and $2$ .

Factor $2$ out of $6$ .

$x = - \frac{2 \cdot 3}{2 \cdot 2}$

Cancel the common factors.

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

$x = - \frac{2 \cdot 3}{2 (2)}$

Cancel the common factor.

$x = - \frac{2 \cdot 3}{2 \cdot 2}$

Rewrite the expression.

$x = - \frac{3}{2}$

Step 3

Evaluate $f (- \frac{3}{2})$ .

Replace the variable $x$ with $- \frac{3}{2}$ in the expression.

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

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 $- \frac{3}{2}$ .

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

Apply the product rule to $\frac{3}{2}$ .

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

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

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

Multiply $\frac{3^{2}}{2^{2}}$ by $1$ .

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

Raise $3$ to the power of $2$ .

$f (- \frac{3}{2}) = 2 (\frac{9}{2^{2}}) + 6 (- \frac{3}{2}) + 3$

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

$f (- \frac{3}{2}) = 2 (\frac{9}{4}) + 6 (- \frac{3}{2}) + 3$

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$f (- \frac{3}{2}) = 2 (\frac{9}{2 (2)}) + 6 (- \frac{3}{2}) + 3$

Cancel the common factor.

$f (- \frac{3}{2}) = 2 (\frac{9}{2 \cdot 2}) + 6 (- \frac{3}{2}) + 3$

Rewrite the expression.

$f (- \frac{3}{2}) = \frac{9}{2} + 6 (- \frac{3}{2}) + 3$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$f (- \frac{3}{2}) = \frac{9}{2} + 6 (\frac{- 3}{2}) + 3$

Factor $2$ out of $6$ .

$f (- \frac{3}{2}) = \frac{9}{2} + 2 (3) (\frac{- 3}{2}) + 3$

Cancel the common factor.

$f (- \frac{3}{2}) = \frac{9}{2} + 2 \cdot (3 (\frac{- 3}{2})) + 3$

Rewrite the expression.

$f (- \frac{3}{2}) = \frac{9}{2} + 3 \cdot - 3 + 3$

Multiply $3$ by $- 3$ .

$f (- \frac{3}{2}) = \frac{9}{2} - 9 + 3$

Find the common denominator.

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

$f (- \frac{3}{2}) = \frac{9}{2} + \frac{- 9}{1} + 3$

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

$f (- \frac{3}{2}) = \frac{9}{2} + \frac{- 9}{1} \cdot \frac{2}{2} + 3$

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

$f (- \frac{3}{2}) = \frac{9}{2} + \frac{- 9 \cdot 2}{2} + 3$

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

$f (- \frac{3}{2}) = \frac{9}{2} + \frac{- 9 \cdot 2}{2} + \frac{3}{1}$

Multiply $\frac{3}{1}$ by $\frac{2}{2}$ .

$f (- \frac{3}{2}) = \frac{9}{2} + \frac{- 9 \cdot 2}{2} + \frac{3}{1} \cdot \frac{2}{2}$

Multiply $\frac{3}{1}$ by $\frac{2}{2}$ .

$f (- \frac{3}{2}) = \frac{9}{2} + \frac{- 9 \cdot 2}{2} + \frac{3 \cdot 2}{2}$

Combine the numerators over the common denominator.

$f (- \frac{3}{2}) = \frac{9 - 9 \cdot 2 + 3 \cdot 2}{2}$

Simplify each term.

Multiply $- 9$ by $2$ .

$f (- \frac{3}{2}) = \frac{9 - 18 + 3 \cdot 2}{2}$

Multiply $3$ by $2$ .

$f (- \frac{3}{2}) = \frac{9 - 18 + 6}{2}$

Simplify the expression.

Subtract $18$ from $9$ .

$f (- \frac{3}{2}) = \frac{- 9 + 6}{2}$

Add $- 9$ and $6$ .

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

Move the negative in front of the fraction.

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

The final answer is $- \frac{3}{2}$ .

$- \frac{3}{2}$

Step 4

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

$(- \frac{3}{2}, - \frac{3}{2})$

Step 5

Find the Maximum/Minimum Value 2x^2+6x+3

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