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.

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

Multiply $2$ 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.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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

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

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

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

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

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

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

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

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

Multiply $- 1$ by $3$ .

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

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 the expression.

Subtract $18$ from $9$ .

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

Subtract $16$ from $- 9$ .

$f (- \frac{3}{2}) = \frac{- 25}{4}$

Move the negative in front of the fraction.

$f (- \frac{3}{2}) = - \frac{25}{4}$

The final answer is $- \frac{25}{4}$ .

$- \frac{25}{4}$

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

$(- \frac{3}{2}, - \frac{25}{4})$

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

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