Find the Maximum/Minimum Value f(x)=-4x^2-6x+1

$f (x) = - 4 x^{2} - 6 x + 1$

The maximum or minimum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is negative, the maximum value of the function is $f (- \frac{b}{2 a})$ . If $a$ is positive, the minimum value of the function is $f (- \frac{b}{2 a})$ .

$f_{max}$ $x = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$

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

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

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

$x = - \frac{- 6}{2 (- 4)}$

Remove parentheses.

$x = - \frac{- 6}{2 (- 4)}$

Simplify $- \frac{- 6}{2 (- 4)}$ .

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Cancel the common factor of $- 6$ and $2$ .

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Factor $2$ out of $- 6$ .

$x = - \frac{2 \cdot - 3}{2 \cdot - 4}$

Cancel the common factors.

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

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

Cancel the common factor.

$x = - \frac{2 \cdot - 3}{2 \cdot - 4}$

Rewrite the expression.

$x = - \frac{- 3}{- 4}$

Dividing two negative values results in a positive value.

$x = - \frac{3}{4}$

Replace the variable $x$ with $- \frac{3}{4}$ in the expression.

$f (- \frac{3}{4}) = - 4 {(- \frac{3}{4})}^{2} - 6 (- \frac{3}{4}) + 1$

Simplify the result.

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Simplify each term.

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

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Apply the product rule to $- \frac{3}{4}$ .

$f (- \frac{3}{4}) = - 4 ({(- 1)}^{2} {(\frac{3}{4})}^{2}) - 6 (- \frac{3}{4}) + 1$

Apply the product rule to $\frac{3}{4}$ .

$f (- \frac{3}{4}) = - 4 ({(- 1)}^{2} (\frac{3^{2}}{4^{2}})) - 6 (- \frac{3}{4}) + 1$

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

$f (- \frac{3}{4}) = - 4 (1 (\frac{3^{2}}{4^{2}})) - 6 (- \frac{3}{4}) + 1$

Multiply $\frac{3^{2}}{4^{2}}$ by $1$ .

$f (- \frac{3}{4}) = - 4 \frac{3^{2}}{4^{2}} - 6 (- \frac{3}{4}) + 1$

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

$f (- \frac{3}{4}) = - 4 \frac{9}{4^{2}} - 6 (- \frac{3}{4}) + 1$

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

$f (- \frac{3}{4}) = - 4 (\frac{9}{16}) - 6 (- \frac{3}{4}) + 1$

Cancel the common factor of $4$ .

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Factor $4$ out of $- 4$ .

$f (- \frac{3}{4}) = 4 (- 1) (\frac{9}{16}) - 6 (- \frac{3}{4}) + 1$

Factor $4$ out of $16$ .

$f (- \frac{3}{4}) = 4 \cdot (- 1 \frac{9}{4 \cdot 4}) - 6 (- \frac{3}{4}) + 1$

Cancel the common factor.

$f (- \frac{3}{4}) = 4 \cdot (- 1 \frac{9}{4 \cdot 4}) - 6 (- \frac{3}{4}) + 1$

Rewrite the expression.

$f (- \frac{3}{4}) = - 1 (\frac{9}{4}) - 6 (- \frac{3}{4}) + 1$

Rewrite $- 1 (\frac{9}{4})$ as $- (\frac{9}{4})$ .

$f (- \frac{3}{4}) = - (\frac{9}{4}) - 6 (- \frac{3}{4}) + 1$

Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{3}{4}$ into the numerator.

$f (- \frac{3}{4}) = - \frac{9}{4} - 6 (\frac{- 3}{4}) + 1$

Factor $2$ out of $- 6$ .

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

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Multiply $- 3$ by $- 3$ .

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

Find the common denominator.

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

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

Combine.

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

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

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

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

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

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

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

Multiply $2$ by $2$ .

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

Combine fractions.

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Combine fractions with similar denominators.

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

Multiply $9$ by $2$ .

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

Simplify the numerator.

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Add $- 9$ and $18$ .

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

Add $9$ and $4$ .

$f (- \frac{3}{4}) = \frac{13}{4}$

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

$\frac{13}{4}$

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

$(- \frac{3}{4}, \frac{13}{4})$

Find the Maximum/Minimum Value f(x)=-4x^2-6x+1

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