Find the Maximum/Minimum Value y=4x^2+x-14

$y = 4 x^{2} + x - 14$

Step 1

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

Step 2

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

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

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

Remove parentheses.

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

Multiply $2$ by $4$ .

$x = - \frac{1}{8}$

Step 3

Evaluate $f (- \frac{1}{8})$ .

Replace the variable $x$ with $- \frac{1}{8}$ in the expression.

$f (- \frac{1}{8}) = 4 {(- \frac{1}{8})}^{2} - \frac{1}{8} - 14$

Simplify the result.

Remove parentheses.

$f (- \frac{1}{8}) = 4 {(- \frac{1}{8})}^{2} - \frac{1}{8} - 14$

Simplify each term.

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

Apply the product rule to $- \frac{1}{8}$ .

$f (- \frac{1}{8}) = 4 ({(- 1)}^{2} {(\frac{1}{8})}^{2}) - \frac{1}{8} - 14$

Apply the product rule to $\frac{1}{8}$ .

$f (- \frac{1}{8}) = 4 ({(- 1)}^{2} (\frac{1^{2}}{8^{2}})) - \frac{1}{8} - 14$

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

$f (- \frac{1}{8}) = 4 (1 (\frac{1^{2}}{8^{2}})) - \frac{1}{8} - 14$

Multiply $\frac{1^{2}}{8^{2}}$ by $1$ .

$f (- \frac{1}{8}) = 4 (\frac{1^{2}}{8^{2}}) - \frac{1}{8} - 14$

One to any power is one.

$f (- \frac{1}{8}) = 4 (\frac{1}{8^{2}}) - \frac{1}{8} - 14$

Raise $8$ to the power of $2$ .

$f (- \frac{1}{8}) = 4 (\frac{1}{64}) - \frac{1}{8} - 14$

Cancel the common factor of $4$ .

Factor $4$ out of $64$ .

$f (- \frac{1}{8}) = 4 (\frac{1}{4 (16)}) - \frac{1}{8} - 14$

Cancel the common factor.

$f (- \frac{1}{8}) = 4 (\frac{1}{4 \cdot 16}) - \frac{1}{8} - 14$

Rewrite the expression.

$f (- \frac{1}{8}) = \frac{1}{16} - \frac{1}{8} - 14$

Find the common denominator.

Multiply $\frac{1}{8}$ by $\frac{2}{2}$ .

$f (- \frac{1}{8}) = \frac{1}{16} - (\frac{1}{8} \cdot \frac{2}{2}) - 14$

Multiply $\frac{1}{8}$ by $\frac{2}{2}$ .

$f (- \frac{1}{8}) = \frac{1}{16} - \frac{2}{8 \cdot 2} - 14$

Write $- 14$ as a fraction with denominator $1$ .

$f (- \frac{1}{8}) = \frac{1}{16} - \frac{2}{8 \cdot 2} + \frac{- 14}{1}$

Multiply $\frac{- 14}{1}$ by $\frac{16}{16}$ .

$f (- \frac{1}{8}) = \frac{1}{16} - \frac{2}{8 \cdot 2} + \frac{- 14}{1} \cdot \frac{16}{16}$

Multiply $\frac{- 14}{1}$ by $\frac{16}{16}$ .

$f (- \frac{1}{8}) = \frac{1}{16} - \frac{2}{8 \cdot 2} + \frac{- 14 \cdot 16}{16}$

Reorder the factors of $8 \cdot 2$ .

$f (- \frac{1}{8}) = \frac{1}{16} - \frac{2}{2 \cdot 8} + \frac{- 14 \cdot 16}{16}$

Multiply $2$ by $8$ .

$f (- \frac{1}{8}) = \frac{1}{16} - \frac{2}{16} + \frac{- 14 \cdot 16}{16}$

Combine the numerators over the common denominator.

$f (- \frac{1}{8}) = \frac{1 - 2 - 14 \cdot 16}{16}$

Simplify the expression.

Multiply $- 14$ by $16$ .

$f (- \frac{1}{8}) = \frac{1 - 2 - 224}{16}$

Subtract $2$ from $1$ .

$f (- \frac{1}{8}) = \frac{- 1 - 224}{16}$

Subtract $224$ from $- 1$ .

$f (- \frac{1}{8}) = \frac{- 225}{16}$

Move the negative in front of the fraction.

$f (- \frac{1}{8}) = - \frac{225}{16}$

The final answer is $- \frac{225}{16}$ .

$- \frac{225}{16}$

Step 4

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

$(- \frac{1}{8}, - \frac{225}{16})$

Step 5

Find the Maximum/Minimum Value y=4x^2+x-14

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