Find the Maximum/Minimum Value x^2-3x+4

$x^{2} - 3 x + 4$

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 $x = - \frac{b}{2 a}$

Find the value of $x = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

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

Remove parentheses.

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

Simplify $- \frac{- 3}{2 (1)}$ .

Multiply $2$ by $1$ .

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

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 $1$ .

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

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

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

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

Simplify the result.

Simplify each term.

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

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

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

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

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

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

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

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

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

Multiply $- 3$ by $3$ .

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

Move the negative in front of the fraction.

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

Find the common denominator.

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

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

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

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

Write $4$ as a fraction with denominator $1$ .

$f (\frac{3}{2}) = \frac{9}{4} - \frac{9 \cdot 2}{2 \cdot 2} + \frac{4}{1}$

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

$f (\frac{3}{2}) = \frac{9}{4} - \frac{9 \cdot 2}{2 \cdot 2} + \frac{4}{1} \cdot \frac{4}{4}$

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

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

Multiply $2$ by $2$ .

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

Combine the numerators over the common denominator.

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

Simplify each term.

Multiply $- 9$ by $2$ .

$f (\frac{3}{2}) = \frac{9 - 18 + 4 \cdot 4}{4}$

Multiply $4$ by $4$ .

$f (\frac{3}{2}) = \frac{9 - 18 + 16}{4}$

Simplify by adding and subtracting.

Subtract $18$ from $9$ .

$f (\frac{3}{2}) = \frac{- 9 + 16}{4}$

Add $- 9$ and $16$ .

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

The final answer is $\frac{7}{4}$ .

$\frac{7}{4}$

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

$(\frac{3}{2}, \frac{7}{4})$

Find the Maximum/Minimum Value x^2-3x+4

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