Find the Maximum/Minimum Value -16x^2+42x+12

$- 16 x^{2} + 42 x + 12$

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

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

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

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

$x = - \frac{42}{2 (- 16)}$

Remove parentheses.

$x = - \frac{42}{2 (- 16)}$

Simplify $- \frac{42}{2 (- 16)}$ .

Cancel the common factor of $42$ and $2$ .

Factor $2$ out of $42$ .

$x = - \frac{2 \cdot 21}{2 \cdot - 16}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 16$ .

$x = - \frac{2 \cdot 21}{2 (- 16)}$

Cancel the common factor.

$x = - \frac{2 \cdot 21}{2 \cdot - 16}$

Rewrite the expression.

$x = - \frac{21}{- 16}$

Move the negative in front of the fraction.

$x = - - \frac{21}{16}$

Multiply $- - \frac{21}{16}$ .

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{21}{16})$

Multiply $\frac{21}{16}$ by $1$ .

$x = \frac{21}{16}$

Evaluate $f (\frac{21}{16})$ .

Replace the variable $x$ with $\frac{21}{16}$ in the expression.

$f (\frac{21}{16}) = - 16 {(\frac{21}{16})}^{2} + 42 (\frac{21}{16}) + 12$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{21}{16}$ .

$f (\frac{21}{16}) = - 16 \frac{21^{2}}{16^{2}} + 42 (\frac{21}{16}) + 12$

Raise $21$ to the power of $2$ .

$f (\frac{21}{16}) = - 16 \frac{441}{16^{2}} + 42 (\frac{21}{16}) + 12$

Raise $16$ to the power of $2$ .

$f (\frac{21}{16}) = - 16 (\frac{441}{256}) + 42 (\frac{21}{16}) + 12$

Cancel the common factor of $16$ .

Factor $16$ out of $- 16$ .

$f (\frac{21}{16}) = 16 (- 1) (\frac{441}{256}) + 42 (\frac{21}{16}) + 12$

Factor $16$ out of $256$ .

$f (\frac{21}{16}) = 16 \cdot (- 1 \frac{441}{16 \cdot 16}) + 42 (\frac{21}{16}) + 12$

Cancel the common factor.

$f (\frac{21}{16}) = 16 \cdot (- 1 \frac{441}{16 \cdot 16}) + 42 (\frac{21}{16}) + 12$

Rewrite the expression.

$f (\frac{21}{16}) = - 1 (\frac{441}{16}) + 42 (\frac{21}{16}) + 12$

Rewrite $- 1 (\frac{441}{16})$ as $- (\frac{441}{16})$ .

$f (\frac{21}{16}) = - (\frac{441}{16}) + 42 (\frac{21}{16}) + 12$

Cancel the common factor of $2$ .

Factor $2$ out of $42$ .

$f (\frac{21}{16}) = - \frac{441}{16} + 2 (21) (\frac{21}{16}) + 12$

Factor $2$ out of $16$ .

$f (\frac{21}{16}) = - \frac{441}{16} + 2 \cdot (21 (\frac{21}{2 \cdot 8})) + 12$

Cancel the common factor.

$f (\frac{21}{16}) = - \frac{441}{16} + 2 \cdot (21 (\frac{21}{2 \cdot 8})) + 12$

Rewrite the expression.

$f (\frac{21}{16}) = - \frac{441}{16} + 21 (\frac{21}{8}) + 12$

Combine $21$ and $\frac{21}{8}$ .

$f (\frac{21}{16}) = - \frac{441}{16} + \frac{21 \cdot 21}{8} + 12$

Multiply $21$ by $21$ .

$f (\frac{21}{16}) = - \frac{441}{16} + \frac{441}{8} + 12$

Find the common denominator.

Multiply $\frac{441}{8}$ by $\frac{2}{2}$ .

$f (\frac{21}{16}) = - \frac{441}{16} + \frac{441}{8} \cdot \frac{2}{2} + 12$

Multiply $\frac{441}{8}$ by $\frac{2}{2}$ .

$f (\frac{21}{16}) = - \frac{441}{16} + \frac{441 \cdot 2}{8 \cdot 2} + 12$

Write $12$ as a fraction with denominator $1$ .

$f (\frac{21}{16}) = - \frac{441}{16} + \frac{441 \cdot 2}{8 \cdot 2} + \frac{12}{1}$

Multiply $\frac{12}{1}$ by $\frac{16}{16}$ .

$f (\frac{21}{16}) = - \frac{441}{16} + \frac{441 \cdot 2}{8 \cdot 2} + \frac{12}{1} \cdot \frac{16}{16}$

Multiply $\frac{12}{1}$ by $\frac{16}{16}$ .

$f (\frac{21}{16}) = - \frac{441}{16} + \frac{441 \cdot 2}{8 \cdot 2} + \frac{12 \cdot 16}{16}$

Reorder the factors of $8 \cdot 2$ .

$f (\frac{21}{16}) = - \frac{441}{16} + \frac{441 \cdot 2}{2 \cdot 8} + \frac{12 \cdot 16}{16}$

Multiply $2$ by $8$ .

$f (\frac{21}{16}) = - \frac{441}{16} + \frac{441 \cdot 2}{16} + \frac{12 \cdot 16}{16}$

Combine the numerators over the common denominator.

$f (\frac{21}{16}) = \frac{- 441 + 441 \cdot 2 + 12 \cdot 16}{16}$

Simplify each term.

Multiply $441$ by $2$ .

$f (\frac{21}{16}) = \frac{- 441 + 882 + 12 \cdot 16}{16}$

Multiply $12$ by $16$ .

$f (\frac{21}{16}) = \frac{- 441 + 882 + 192}{16}$

Simplify by adding numbers.

Add $- 441$ and $882$ .

$f (\frac{21}{16}) = \frac{441 + 192}{16}$

Add $441$ and $192$ .

$f (\frac{21}{16}) = \frac{633}{16}$

The final answer is $\frac{633}{16}$ .

$\frac{633}{16}$

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

$(\frac{21}{16}, \frac{633}{16})$

Find the Maximum/Minimum Value -16x^2+42x+12

Related Questions

Support LCMGCF.com by disabling your adblocker.