Find the Maximum/Minimum Value f(x)=2x^2-6x+7

$f (x) = 2 x^{2} - 6 x + 7$

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.

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

Simplify $- \frac{- 6}{2 (2)}$ .

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.

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

Move the negative in front of the fraction.

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

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

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{3}{2})$

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

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

Step 3

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

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

$f (\frac{3}{2}) = 2 {(\frac{3}{2})}^{2} - 6 (\frac{3}{2}) + 7$

Simplify the result.

Simplify each term.

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

$f (\frac{3}{2}) = 2 (\frac{3^{2}}{2^{2}}) - 6 (\frac{3}{2}) + 7$

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

$f (\frac{3}{2}) = 2 (\frac{9}{2^{2}}) - 6 (\frac{3}{2}) + 7$

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

$f (\frac{3}{2}) = 2 (\frac{9}{4}) - 6 (\frac{3}{2}) + 7$

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$f (\frac{3}{2}) = 2 (\frac{9}{2 (2)}) - 6 (\frac{3}{2}) + 7$

Cancel the common factor.

$f (\frac{3}{2}) = 2 (\frac{9}{2 \cdot 2}) - 6 (\frac{3}{2}) + 7$

Rewrite the expression.

$f (\frac{3}{2}) = \frac{9}{2} - 6 (\frac{3}{2}) + 7$

Cancel the common factor of $2$ .

Factor $2$ out of $- 6$ .

$f (\frac{3}{2}) = \frac{9}{2} + 2 (- 3) (\frac{3}{2}) + 7$

Cancel the common factor.

$f (\frac{3}{2}) = \frac{9}{2} + 2 \cdot (- 3 (\frac{3}{2})) + 7$

Rewrite the expression.

$f (\frac{3}{2}) = \frac{9}{2} - 3 \cdot 3 + 7$

Multiply $- 3$ by $3$ .

$f (\frac{3}{2}) = \frac{9}{2} - 9 + 7$

Find the common denominator.

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

$f (\frac{3}{2}) = \frac{9}{2} + \frac{- 9}{1} + 7$

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

$f (\frac{3}{2}) = \frac{9}{2} + \frac{- 9}{1} \cdot \frac{2}{2} + 7$

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

$f (\frac{3}{2}) = \frac{9}{2} + \frac{- 9 \cdot 2}{2} + 7$

Write $7$ as a fraction with denominator $1$ .

$f (\frac{3}{2}) = \frac{9}{2} + \frac{- 9 \cdot 2}{2} + \frac{7}{1}$

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

$f (\frac{3}{2}) = \frac{9}{2} + \frac{- 9 \cdot 2}{2} + \frac{7}{1} \cdot \frac{2}{2}$

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

$f (\frac{3}{2}) = \frac{9}{2} + \frac{- 9 \cdot 2}{2} + \frac{7 \cdot 2}{2}$

Combine the numerators over the common denominator.

$f (\frac{3}{2}) = \frac{9 - 9 \cdot 2 + 7 \cdot 2}{2}$

Simplify each term.

Multiply $- 9$ by $2$ .

$f (\frac{3}{2}) = \frac{9 - 18 + 7 \cdot 2}{2}$

Multiply $7$ by $2$ .

$f (\frac{3}{2}) = \frac{9 - 18 + 14}{2}$

Simplify by adding and subtracting.

Subtract $18$ from $9$ .

$f (\frac{3}{2}) = \frac{- 9 + 14}{2}$

Add $- 9$ and $14$ .

$f (\frac{3}{2}) = \frac{5}{2}$

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

$\frac{5}{2}$

Step 4

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

$(\frac{3}{2}, \frac{5}{2})$

Step 5

Find the Maximum/Minimum Value f(x)=2x^2-6x+7

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