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

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

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 $b$ .

$x = - \frac{- 2}{2 (4)}$

Remove parentheses.

$x = - \frac{- 2}{2 (4)}$

Simplify $- \frac{- 2}{2 (4)}$ .

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

Factor $2$ out of $- 2$ .

$x = - \frac{2 \cdot - 1}{2 \cdot 4}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 4$ .

$x = - \frac{2 \cdot - 1}{2 (4)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 1}{2 \cdot 4}$

Rewrite the expression.

$x = - \frac{- 1}{4}$

Move the negative in front of the fraction.

$x = - - \frac{1}{4}$

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

Multiply $- 1$ by $- 1$ .

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

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

$x = \frac{1}{4}$

Evaluate $f (\frac{1}{4})$ .

Replace the variable $x$ with $\frac{1}{4}$ in the expression.

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

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{1}{4}$ .

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

One to any power is one.

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

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

$f (\frac{1}{4}) = 4 (\frac{1}{16}) - 2 (\frac{1}{4}) - 3$

Cancel the common factor of $4$ .

Factor $4$ out of $16$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

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

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Rewrite $- 1 (\frac{1}{2})$ as $- (\frac{1}{2})$ .

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

Find the common denominator.

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

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

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

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

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

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

Multiply $\frac{- 3}{1}$ by $\frac{4}{4}$ .

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

Multiply $\frac{- 3}{1}$ by $\frac{4}{4}$ .

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

Multiply $2$ by $2$ .

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

Combine the numerators over the common denominator.

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

Simplify the expression.

Multiply $- 3$ by $4$ .

$f (\frac{1}{4}) = \frac{1 - 2 - 12}{4}$

Subtract $2$ from $1$ .

$f (\frac{1}{4}) = \frac{- 1 - 12}{4}$

Subtract $12$ from $- 1$ .

$f (\frac{1}{4}) = \frac{- 13}{4}$

Move the negative in front of the fraction.

$f (\frac{1}{4}) = - \frac{13}{4}$

The final answer is $- \frac{13}{4}$ .

$- \frac{13}{4}$

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

$(\frac{1}{4}, - \frac{13}{4})$

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

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