Find the Maximum/Minimum Value f(x)=-16x^2+22x+3

$f (x) = - 16 x^{2} + 22 x + 3$

Step 1

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

Step 2

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

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

$x = - \frac{22}{2 (- 16)}$

Remove parentheses.

$x = - \frac{22}{2 (- 16)}$

Simplify $- \frac{22}{2 (- 16)}$ .

Cancel the common factor of $22$ and $2$ .

Factor $2$ out of $22$ .

$x = - \frac{2 \cdot 11}{2 \cdot - 16}$

Cancel the common factors.

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

$x = - \frac{2 \cdot 11}{2 (- 16)}$

Cancel the common factor.

$x = - \frac{2 \cdot 11}{2 \cdot - 16}$

Rewrite the expression.

$x = - \frac{11}{- 16}$

Move the negative in front of the fraction.

$x = - - \frac{11}{16}$

Multiply $- - \frac{11}{16}$ .

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{11}{16})$

Multiply $\frac{11}{16}$ by $1$ .

$x = \frac{11}{16}$

Step 3

Evaluate $f (\frac{11}{16})$ .

Replace the variable $x$ with $\frac{11}{16}$ in the expression.

$f (\frac{11}{16}) = - 16 {(\frac{11}{16})}^{2} + 22 (\frac{11}{16}) + 3$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{11}{16}$ .

$f (\frac{11}{16}) = - 16 \frac{11^{2}}{16^{2}} + 22 (\frac{11}{16}) + 3$

Raise $11$ to the power of $2$ .

$f (\frac{11}{16}) = - 16 \frac{121}{16^{2}} + 22 (\frac{11}{16}) + 3$

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

$f (\frac{11}{16}) = - 16 (\frac{121}{256}) + 22 (\frac{11}{16}) + 3$

Cancel the common factor of $16$ .

Factor $16$ out of $- 16$ .

$f (\frac{11}{16}) = 16 (- 1) (\frac{121}{256}) + 22 (\frac{11}{16}) + 3$

Factor $16$ out of $256$ .

$f (\frac{11}{16}) = 16 \cdot (- 1 \frac{121}{16 \cdot 16}) + 22 (\frac{11}{16}) + 3$

Cancel the common factor.

$f (\frac{11}{16}) = 16 \cdot (- 1 \frac{121}{16 \cdot 16}) + 22 (\frac{11}{16}) + 3$

Rewrite the expression.

$f (\frac{11}{16}) = - 1 (\frac{121}{16}) + 22 (\frac{11}{16}) + 3$

Rewrite $- 1 (\frac{121}{16})$ as $- (\frac{121}{16})$ .

$f (\frac{11}{16}) = - (\frac{121}{16}) + 22 (\frac{11}{16}) + 3$

Cancel the common factor of $2$ .

Factor $2$ out of $22$ .

$f (\frac{11}{16}) = - \frac{121}{16} + 2 (11) (\frac{11}{16}) + 3$

Factor $2$ out of $16$ .

$f (\frac{11}{16}) = - \frac{121}{16} + 2 \cdot (11 (\frac{11}{2 \cdot 8})) + 3$

Cancel the common factor.

$f (\frac{11}{16}) = - \frac{121}{16} + 2 \cdot (11 (\frac{11}{2 \cdot 8})) + 3$

Rewrite the expression.

$f (\frac{11}{16}) = - \frac{121}{16} + 11 (\frac{11}{8}) + 3$

Combine $11$ and $\frac{11}{8}$ .

$f (\frac{11}{16}) = - \frac{121}{16} + \frac{11 \cdot 11}{8} + 3$

Multiply $11$ by $11$ .

$f (\frac{11}{16}) = - \frac{121}{16} + \frac{121}{8} + 3$

Find the common denominator.

Multiply $\frac{121}{8}$ by $\frac{2}{2}$ .

$f (\frac{11}{16}) = - \frac{121}{16} + \frac{121}{8} \cdot \frac{2}{2} + 3$

Multiply $\frac{121}{8}$ by $\frac{2}{2}$ .

$f (\frac{11}{16}) = - \frac{121}{16} + \frac{121 \cdot 2}{8 \cdot 2} + 3$

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

$f (\frac{11}{16}) = - \frac{121}{16} + \frac{121 \cdot 2}{8 \cdot 2} + \frac{3}{1}$

Multiply $\frac{3}{1}$ by $\frac{16}{16}$ .

$f (\frac{11}{16}) = - \frac{121}{16} + \frac{121 \cdot 2}{8 \cdot 2} + \frac{3}{1} \cdot \frac{16}{16}$

Multiply $\frac{3}{1}$ by $\frac{16}{16}$ .

$f (\frac{11}{16}) = - \frac{121}{16} + \frac{121 \cdot 2}{8 \cdot 2} + \frac{3 \cdot 16}{16}$

Reorder the factors of $8 \cdot 2$ .

$f (\frac{11}{16}) = - \frac{121}{16} + \frac{121 \cdot 2}{2 \cdot 8} + \frac{3 \cdot 16}{16}$

Multiply $2$ by $8$ .

$f (\frac{11}{16}) = - \frac{121}{16} + \frac{121 \cdot 2}{16} + \frac{3 \cdot 16}{16}$

Combine the numerators over the common denominator.

$f (\frac{11}{16}) = \frac{- 121 + 121 \cdot 2 + 3 \cdot 16}{16}$

Simplify each term.

Multiply $121$ by $2$ .

$f (\frac{11}{16}) = \frac{- 121 + 242 + 3 \cdot 16}{16}$

Multiply $3$ by $16$ .

$f (\frac{11}{16}) = \frac{- 121 + 242 + 48}{16}$

Simplify by adding numbers.

Add $- 121$ and $242$ .

$f (\frac{11}{16}) = \frac{121 + 48}{16}$

Add $121$ and $48$ .

$f (\frac{11}{16}) = \frac{169}{16}$

The final answer is $\frac{169}{16}$ .

$\frac{169}{16}$

Step 4

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

$(\frac{11}{16}, \frac{169}{16})$

Step 5

Find the Maximum/Minimum Value f(x)=-16x^2+22x+3

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