Find the Maximum/Minimum Value f(x)=2x^2+14x+27

$f (x) = 2 x^{2} + 14 x + 27$

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 $b$ .

$x = - \frac{14}{2 (2)}$

Remove parentheses.

$x = - \frac{14}{2 (2)}$

Cancel the common factor of $14$ and $2$ .

Factor $2$ out of $14$ .

$x = - \frac{2 \cdot 7}{2 \cdot 2}$

Cancel the common factors.

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

$x = - \frac{2 \cdot 7}{2 (2)}$

Cancel the common factor.

$x = - \frac{2 \cdot 7}{2 \cdot 2}$

Rewrite the expression.

$x = - \frac{7}{2}$

Step 3

Evaluate $f (- \frac{7}{2})$ .

Replace the variable $x$ with $- \frac{7}{2}$ in the expression.

$f (- \frac{7}{2}) = 2 {(- \frac{7}{2})}^{2} + 14 (- \frac{7}{2}) + 27$

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 $- \frac{7}{2}$ .

$f (- \frac{7}{2}) = 2 ({(- 1)}^{2} {(\frac{7}{2})}^{2}) + 14 (- \frac{7}{2}) + 27$

Apply the product rule to $\frac{7}{2}$ .

$f (- \frac{7}{2}) = 2 ({(- 1)}^{2} (\frac{7^{2}}{2^{2}})) + 14 (- \frac{7}{2}) + 27$

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

$f (- \frac{7}{2}) = 2 (1 (\frac{7^{2}}{2^{2}})) + 14 (- \frac{7}{2}) + 27$

Multiply $\frac{7^{2}}{2^{2}}$ by $1$ .

$f (- \frac{7}{2}) = 2 (\frac{7^{2}}{2^{2}}) + 14 (- \frac{7}{2}) + 27$

Raise $7$ to the power of $2$ .

$f (- \frac{7}{2}) = 2 (\frac{49}{2^{2}}) + 14 (- \frac{7}{2}) + 27$

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

$f (- \frac{7}{2}) = 2 (\frac{49}{4}) + 14 (- \frac{7}{2}) + 27$

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$f (- \frac{7}{2}) = 2 (\frac{49}{2 (2)}) + 14 (- \frac{7}{2}) + 27$

Cancel the common factor.

$f (- \frac{7}{2}) = 2 (\frac{49}{2 \cdot 2}) + 14 (- \frac{7}{2}) + 27$

Rewrite the expression.

$f (- \frac{7}{2}) = \frac{49}{2} + 14 (- \frac{7}{2}) + 27$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{7}{2}$ into the numerator.

$f (- \frac{7}{2}) = \frac{49}{2} + 14 (\frac{- 7}{2}) + 27$

Factor $2$ out of $14$ .

$f (- \frac{7}{2}) = \frac{49}{2} + 2 (7) (\frac{- 7}{2}) + 27$

Cancel the common factor.

$f (- \frac{7}{2}) = \frac{49}{2} + 2 \cdot (7 (\frac{- 7}{2})) + 27$

Rewrite the expression.

$f (- \frac{7}{2}) = \frac{49}{2} + 7 \cdot - 7 + 27$

Multiply $7$ by $- 7$ .

$f (- \frac{7}{2}) = \frac{49}{2} - 49 + 27$

Find the common denominator.

Write $- 49$ as a fraction with denominator $1$ .

$f (- \frac{7}{2}) = \frac{49}{2} + \frac{- 49}{1} + 27$

Multiply $\frac{- 49}{1}$ by $\frac{2}{2}$ .

$f (- \frac{7}{2}) = \frac{49}{2} + \frac{- 49}{1} \cdot \frac{2}{2} + 27$

Multiply $\frac{- 49}{1}$ by $\frac{2}{2}$ .

$f (- \frac{7}{2}) = \frac{49}{2} + \frac{- 49 \cdot 2}{2} + 27$

Write $27$ as a fraction with denominator $1$ .

$f (- \frac{7}{2}) = \frac{49}{2} + \frac{- 49 \cdot 2}{2} + \frac{27}{1}$

Multiply $\frac{27}{1}$ by $\frac{2}{2}$ .

$f (- \frac{7}{2}) = \frac{49}{2} + \frac{- 49 \cdot 2}{2} + \frac{27}{1} \cdot \frac{2}{2}$

Multiply $\frac{27}{1}$ by $\frac{2}{2}$ .

$f (- \frac{7}{2}) = \frac{49}{2} + \frac{- 49 \cdot 2}{2} + \frac{27 \cdot 2}{2}$

Combine the numerators over the common denominator.

$f (- \frac{7}{2}) = \frac{49 - 49 \cdot 2 + 27 \cdot 2}{2}$

Simplify each term.

Multiply $- 49$ by $2$ .

$f (- \frac{7}{2}) = \frac{49 - 98 + 27 \cdot 2}{2}$

Multiply $27$ by $2$ .

$f (- \frac{7}{2}) = \frac{49 - 98 + 54}{2}$

Simplify by adding and subtracting.

Subtract $98$ from $49$ .

$f (- \frac{7}{2}) = \frac{- 49 + 54}{2}$

Add $- 49$ and $54$ .

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

The final answer is $\frac{5}{2}$ .

$\frac{5}{2}$

Step 4

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

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

Step 5

Find the Maximum/Minimum Value f(x)=2x^2+14x+27

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