Find the Maximum/Minimum Value -6x^2+10x+1

$- 6 x^{2} + 10 x + 1$

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 $b$ .

$x = - \frac{10}{2 (- 6)}$

Remove parentheses.

$x = - \frac{10}{2 (- 6)}$

Simplify $- \frac{10}{2 (- 6)}$ .

Cancel the common factor of $10$ and $2$ .

Factor $2$ out of $10$ .

$x = - \frac{2 \cdot 5}{2 \cdot - 6}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 6$ .

$x = - \frac{2 \cdot 5}{2 (- 6)}$

Cancel the common factor.

$x = - \frac{2 \cdot 5}{2 \cdot - 6}$

Rewrite the expression.

$x = - \frac{5}{- 6}$

Move the negative in front of the fraction.

$x = - - \frac{5}{6}$

Multiply $- - \frac{5}{6}$ .

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{5}{6})$

Multiply $\frac{5}{6}$ by $1$ .

$x = \frac{5}{6}$

Step 3

Evaluate $f (\frac{5}{6})$ .

Replace the variable $x$ with $\frac{5}{6}$ in the expression.

$f (\frac{5}{6}) = - 6 {(\frac{5}{6})}^{2} + 10 (\frac{5}{6}) + 1$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{5}{6}$ .

$f (\frac{5}{6}) = - 6 \frac{5^{2}}{6^{2}} + 10 (\frac{5}{6}) + 1$

Raise $5$ to the power of $2$ .

$f (\frac{5}{6}) = - 6 \frac{25}{6^{2}} + 10 (\frac{5}{6}) + 1$

Raise $6$ to the power of $2$ .

$f (\frac{5}{6}) = - 6 (\frac{25}{36}) + 10 (\frac{5}{6}) + 1$

Cancel the common factor of $6$ .

Factor $6$ out of $- 6$ .

$f (\frac{5}{6}) = 6 (- 1) (\frac{25}{36}) + 10 (\frac{5}{6}) + 1$

Factor $6$ out of $36$ .

$f (\frac{5}{6}) = 6 \cdot (- 1 \frac{25}{6 \cdot 6}) + 10 (\frac{5}{6}) + 1$

Cancel the common factor.

$f (\frac{5}{6}) = 6 \cdot (- 1 \frac{25}{6 \cdot 6}) + 10 (\frac{5}{6}) + 1$

Rewrite the expression.

$f (\frac{5}{6}) = - 1 (\frac{25}{6}) + 10 (\frac{5}{6}) + 1$

Rewrite $- 1 (\frac{25}{6})$ as $- (\frac{25}{6})$ .

$f (\frac{5}{6}) = - (\frac{25}{6}) + 10 (\frac{5}{6}) + 1$

Cancel the common factor of $2$ .

Factor $2$ out of $10$ .

$f (\frac{5}{6}) = - \frac{25}{6} + 2 (5) (\frac{5}{6}) + 1$

Factor $2$ out of $6$ .

$f (\frac{5}{6}) = - \frac{25}{6} + 2 \cdot (5 (\frac{5}{2 \cdot 3})) + 1$

Cancel the common factor.

$f (\frac{5}{6}) = - \frac{25}{6} + 2 \cdot (5 (\frac{5}{2 \cdot 3})) + 1$

Rewrite the expression.

$f (\frac{5}{6}) = - \frac{25}{6} + 5 (\frac{5}{3}) + 1$

Combine $5$ and $\frac{5}{3}$ .

$f (\frac{5}{6}) = - \frac{25}{6} + \frac{5 \cdot 5}{3} + 1$

Multiply $5$ by $5$ .

$f (\frac{5}{6}) = - \frac{25}{6} + \frac{25}{3} + 1$

Find the common denominator.

Multiply $\frac{25}{3}$ by $\frac{2}{2}$ .

$f (\frac{5}{6}) = - \frac{25}{6} + \frac{25}{3} \cdot \frac{2}{2} + 1$

Multiply $\frac{25}{3}$ by $\frac{2}{2}$ .

$f (\frac{5}{6}) = - \frac{25}{6} + \frac{25 \cdot 2}{3 \cdot 2} + 1$

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

$f (\frac{5}{6}) = - \frac{25}{6} + \frac{25 \cdot 2}{3 \cdot 2} + \frac{1}{1}$

Multiply $\frac{1}{1}$ by $\frac{6}{6}$ .

$f (\frac{5}{6}) = - \frac{25}{6} + \frac{25 \cdot 2}{3 \cdot 2} + \frac{1}{1} \cdot \frac{6}{6}$

Multiply $\frac{1}{1}$ by $\frac{6}{6}$ .

$f (\frac{5}{6}) = - \frac{25}{6} + \frac{25 \cdot 2}{3 \cdot 2} + \frac{6}{6}$

Reorder the factors of $3 \cdot 2$ .

$f (\frac{5}{6}) = - \frac{25}{6} + \frac{25 \cdot 2}{2 \cdot 3} + \frac{6}{6}$

Multiply $2$ by $3$ .

$f (\frac{5}{6}) = - \frac{25}{6} + \frac{25 \cdot 2}{6} + \frac{6}{6}$

Combine the numerators over the common denominator.

$f (\frac{5}{6}) = \frac{- 25 + 25 \cdot 2 + 6}{6}$

Simplify the expression.

Multiply $25$ by $2$ .

$f (\frac{5}{6}) = \frac{- 25 + 50 + 6}{6}$

Add $- 25$ and $50$ .

$f (\frac{5}{6}) = \frac{25 + 6}{6}$

Add $25$ and $6$ .

$f (\frac{5}{6}) = \frac{31}{6}$

The final answer is $\frac{31}{6}$ .

$\frac{31}{6}$

Step 4

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

$(\frac{5}{6}, \frac{31}{6})$

Step 5

Find the Maximum/Minimum Value -6x^2+10x+1

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