Algebra Examples

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

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

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Step 2.1

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

$x = - \frac{6}{2 (- 1)}$

Step 2.2

Remove parentheses.

$x = - \frac{6}{2 (- 1)}$

Step 2.3

Simplify $- \frac{6}{2 (- 1)}$ .

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Step 2.3.1

Cancel the common factor of $6$ and $2$ .

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Step 2.3.1.1

Factor $2$ out of $6$ .

$x = - \frac{2 \cdot 3}{2 \cdot - 1}$

Step 2.3.1.2

Move the negative one from the denominator of $\frac{3}{- 1}$ .

$x = - (- 1 \cdot 3)$

Step 2.3.2

Multiply $- (- 1 \cdot 3)$ .

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Step 2.3.2.1

Multiply $- 1$ by $3$ .

$x = - - 3$

Step 2.3.2.2

Multiply $- 1$ by $- 3$ .

$x = 3$

Step 3

Evaluate $f (3)$ .

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Step 3.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = - {(3)}^{2} + 6 (3) + 1$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Raise $3$ to the power of $2$ .

$f (3) = - 1 \cdot 9 + 6 (3) + 1$

Step 3.2.1.2

Multiply $- 1$ by $9$ .

$f (3) = - 9 + 6 (3) + 1$

Step 3.2.1.3

Multiply $6$ by $3$ .

$f (3) = - 9 + 18 + 1$

Step 3.2.2

Simplify by adding numbers.

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Step 3.2.2.1

Add $- 9$ and $18$ .

$f (3) = 9 + 1$

Step 3.2.2.2

Add $9$ and $1$ .

$f (3) = 10$

Step 3.2.3

The final answer is $10$ .

$10$

Step 4

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

$(3, 10)$

Step 5