Find the Maximum/Minimum Value 4x^2+2x-7

$4 x^{2} + 2 x - 7$

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.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

$f (- \frac{1}{4}) = 4 {(- \frac{1}{4})}^{2} + 2 (- \frac{1}{4}) - 7$

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

$f (- \frac{1}{4}) = 4 ({(- 1)}^{2} {(\frac{1}{4})}^{2}) + 2 (- \frac{1}{4}) - 7$

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

$f (- \frac{1}{4}) = 4 ({(- 1)}^{2} (\frac{1^{2}}{4^{2}})) + 2 (- \frac{1}{4}) - 7$

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

$f (- \frac{1}{4}) = 4 (1 (\frac{1^{2}}{4^{2}})) + 2 (- \frac{1}{4}) - 7$

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

$f (- \frac{1}{4}) = 4 (\frac{1^{2}}{4^{2}}) + 2 (- \frac{1}{4}) - 7$

One to any power is one.

$f (- \frac{1}{4}) = 4 (\frac{1}{4^{2}}) + 2 (- \frac{1}{4}) - 7$

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

$f (- \frac{1}{4}) = 4 (\frac{1}{16}) + 2 (- \frac{1}{4}) - 7$

Cancel the common factor of $4$ .

Factor $4$ out of $16$ .

$f (- \frac{1}{4}) = 4 (\frac{1}{4 (4)}) + 2 (- \frac{1}{4}) - 7$

Cancel the common factor.

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

Rewrite the expression.

$f (- \frac{1}{4}) = \frac{1}{4} + 2 (- \frac{1}{4}) - 7$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{4}$ into the numerator.

$f (- \frac{1}{4}) = \frac{1}{4} + 2 (\frac{- 1}{4}) - 7$

Factor $2$ out of $4$ .

$f (- \frac{1}{4}) = \frac{1}{4} + 2 (\frac{- 1}{2 (2)}) - 7$

Cancel the common factor.

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

Rewrite the expression.

$f (- \frac{1}{4}) = \frac{1}{4} + \frac{- 1}{2} - 7$

Move the negative in front of the fraction.

$f (- \frac{1}{4}) = \frac{1}{4} - \frac{1}{2} - 7$

Find the common denominator.

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

$f (- \frac{1}{4}) = \frac{1}{4} - (\frac{1}{2} \cdot \frac{2}{2}) - 7$

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

$f (- \frac{1}{4}) = \frac{1}{4} - \frac{2}{2 \cdot 2} - 7$

Write $- 7$ as a fraction with denominator $1$ .

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

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

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

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

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

Multiply $2$ by $2$ .

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

Combine the numerators over the common denominator.

$f (- \frac{1}{4}) = \frac{1 - 2 - 7 \cdot 4}{4}$

Simplify the expression.

Multiply $- 7$ by $4$ .

$f (- \frac{1}{4}) = \frac{1 - 2 - 28}{4}$

Subtract $2$ from $1$ .

$f (- \frac{1}{4}) = \frac{- 1 - 28}{4}$

Subtract $28$ from $- 1$ .

$f (- \frac{1}{4}) = \frac{- 29}{4}$

Move the negative in front of the fraction.

$f (- \frac{1}{4}) = - \frac{29}{4}$

The final answer is $- \frac{29}{4}$ .

$- \frac{29}{4}$

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

$(- \frac{1}{4}, - \frac{29}{4})$

Find the Maximum/Minimum Value 4x^2+2x-7

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