Find the Maximum/Minimum Value f(x)=2x^2-7x+3

$f (x) = 2 x^{2} - 7 x + 3$

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 $b$ .

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

Remove parentheses.

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

Simplify $- \frac{- 7}{2 (2)}$ .

Multiply $2$ by $2$ .

$x = - \frac{- 7}{4}$

Move the negative in front of the fraction.

$x = - - \frac{7}{4}$

Multiply $- - \frac{7}{4}$ .

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{7}{4})$

Multiply $\frac{7}{4}$ by $1$ .

$x = \frac{7}{4}$

Step 3

Evaluate $f (\frac{7}{4})$ .

Replace the variable $x$ with $\frac{7}{4}$ in the expression.

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

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{7}{4}$ .

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

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

$f (\frac{7}{4}) = 2 (\frac{49}{4^{2}}) - 7 (\frac{7}{4}) + 3$

Raise $4$ to the power of $2$ .

$f (\frac{7}{4}) = 2 (\frac{49}{16}) - 7 (\frac{7}{4}) + 3$

Cancel the common factor of $2$ .

Factor $2$ out of $16$ .

$f (\frac{7}{4}) = 2 (\frac{49}{2 (8)}) - 7 (\frac{7}{4}) + 3$

Cancel the common factor.

$f (\frac{7}{4}) = 2 (\frac{49}{2 \cdot 8}) - 7 (\frac{7}{4}) + 3$

Rewrite the expression.

$f (\frac{7}{4}) = \frac{49}{8} - 7 (\frac{7}{4}) + 3$

Multiply $- 7 (\frac{7}{4})$ .

Combine $- 7$ and $\frac{7}{4}$ .

$f (\frac{7}{4}) = \frac{49}{8} + \frac{- 7 \cdot 7}{4} + 3$

Multiply $- 7$ by $7$ .

$f (\frac{7}{4}) = \frac{49}{8} + \frac{- 49}{4} + 3$

Move the negative in front of the fraction.

$f (\frac{7}{4}) = \frac{49}{8} - \frac{49}{4} + 3$

Find the common denominator.

Multiply $\frac{49}{4}$ by $\frac{2}{2}$ .

$f (\frac{7}{4}) = \frac{49}{8} - (\frac{49}{4} \cdot \frac{2}{2}) + 3$

Multiply $\frac{49}{4}$ by $\frac{2}{2}$ .

$f (\frac{7}{4}) = \frac{49}{8} - \frac{49 \cdot 2}{4 \cdot 2} + 3$

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

$f (\frac{7}{4}) = \frac{49}{8} - \frac{49 \cdot 2}{4 \cdot 2} + \frac{3}{1}$

Multiply $\frac{3}{1}$ by $\frac{8}{8}$ .

$f (\frac{7}{4}) = \frac{49}{8} - \frac{49 \cdot 2}{4 \cdot 2} + \frac{3}{1} \cdot \frac{8}{8}$

Multiply $\frac{3}{1}$ by $\frac{8}{8}$ .

$f (\frac{7}{4}) = \frac{49}{8} - \frac{49 \cdot 2}{4 \cdot 2} + \frac{3 \cdot 8}{8}$

Reorder the factors of $4 \cdot 2$ .

$f (\frac{7}{4}) = \frac{49}{8} - \frac{49 \cdot 2}{2 \cdot 4} + \frac{3 \cdot 8}{8}$

Multiply $2$ by $4$ .

$f (\frac{7}{4}) = \frac{49}{8} - \frac{49 \cdot 2}{8} + \frac{3 \cdot 8}{8}$

Combine the numerators over the common denominator.

$f (\frac{7}{4}) = \frac{49 - 49 \cdot 2 + 3 \cdot 8}{8}$

Simplify each term.

Multiply $- 49$ by $2$ .

$f (\frac{7}{4}) = \frac{49 - 98 + 3 \cdot 8}{8}$

Multiply $3$ by $8$ .

$f (\frac{7}{4}) = \frac{49 - 98 + 24}{8}$

Simplify the expression.

Subtract $98$ from $49$ .

$f (\frac{7}{4}) = \frac{- 49 + 24}{8}$

Add $- 49$ and $24$ .

$f (\frac{7}{4}) = \frac{- 25}{8}$

Move the negative in front of the fraction.

$f (\frac{7}{4}) = - \frac{25}{8}$

The final answer is $- \frac{25}{8}$ .

$- \frac{25}{8}$

Step 4

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

$(\frac{7}{4}, - \frac{25}{8})$

Step 5

Find the Maximum/Minimum Value f(x)=2x^2-7x+3

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