Find the Maximum/Minimum Value y=-3x^2-2x-5

$y = - 3 x^{2} - 2 x - 5$

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 $b$ .

$x = - \frac{- 2}{2 (- 3)}$

Remove parentheses.

$x = - \frac{- 2}{2 (- 3)}$

Simplify $- \frac{- 2}{2 (- 3)}$ .

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

Factor $2$ out of $- 2$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot - 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Dividing two negative values results in a positive value.

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

Step 3

Evaluate $f (- \frac{1}{3})$ .

Replace the variable $x$ with $- \frac{1}{3}$ in the expression.

$f (- \frac{1}{3}) = - 3 {(- \frac{1}{3})}^{2} - 2 (- \frac{1}{3}) - 5$

Simplify the result.

Simplify each term.

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

Apply the product rule to $- \frac{1}{3}$ .

$f (- \frac{1}{3}) = - 3 ({(- 1)}^{2} {(\frac{1}{3})}^{2}) - 2 (- \frac{1}{3}) - 5$

Apply the product rule to $\frac{1}{3}$ .

$f (- \frac{1}{3}) = - 3 ({(- 1)}^{2} (\frac{1^{2}}{3^{2}})) - 2 (- \frac{1}{3}) - 5$

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

$f (- \frac{1}{3}) = - 3 (1 (\frac{1^{2}}{3^{2}})) - 2 (- \frac{1}{3}) - 5$

Multiply $\frac{1^{2}}{3^{2}}$ by $1$ .

$f (- \frac{1}{3}) = - 3 \frac{1^{2}}{3^{2}} - 2 (- \frac{1}{3}) - 5$

One to any power is one.

$f (- \frac{1}{3}) = - 3 \frac{1}{3^{2}} - 2 (- \frac{1}{3}) - 5$

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

$f (- \frac{1}{3}) = - 3 (\frac{1}{9}) - 2 (- \frac{1}{3}) - 5$

Cancel the common factor of $3$ .

Factor $3$ out of $- 3$ .

$f (- \frac{1}{3}) = 3 (- 1) (\frac{1}{9}) - 2 (- \frac{1}{3}) - 5$

Factor $3$ out of $9$ .

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

Cancel the common factor.

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

Rewrite the expression.

$f (- \frac{1}{3}) = - 1 (\frac{1}{3}) - 2 (- \frac{1}{3}) - 5$

Rewrite $- 1 (\frac{1}{3})$ as $- (\frac{1}{3})$ .

$f (- \frac{1}{3}) = - (\frac{1}{3}) - 2 (- \frac{1}{3}) - 5$

Multiply $- 2 (- \frac{1}{3})$ .

Multiply $- 1$ by $- 2$ .

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

Combine $2$ and $\frac{1}{3}$ .

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

Combine fractions.

Combine the numerators over the common denominator.

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

Add $- 1$ and $2$ .

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

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 5$ by $3$ .

$f (- \frac{1}{3}) = \frac{- 15 + 1}{3}$

Add $- 15$ and $1$ .

$f (- \frac{1}{3}) = \frac{- 14}{3}$

Move the negative in front of the fraction.

$f (- \frac{1}{3}) = - \frac{14}{3}$

The final answer is $- \frac{14}{3}$ .

$- \frac{14}{3}$

Step 4

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

$(- \frac{1}{3}, - \frac{14}{3})$

Step 5

Find the Maximum/Minimum Value y=-3x^2-2x-5

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