Find the Maximum/Minimum Value f(x)=3x^2+2x-9

$f (x) = 3 x^{2} + 2 x - 9$

Step 1

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

$f_{min}$ $x = a x^{2} + b x + c$ 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.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 3

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

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

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

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 $- \frac{1}{3}$ .

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

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

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

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

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

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

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

One to any power is one.

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

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

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

Cancel the common factor of $3$ .

Factor $3$ out of $9$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Multiply $- 1$ by $2$ .

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

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

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

Move the negative in front of the fraction.

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

Combine fractions.

Combine the numerators over the common denominator.

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

Simplify the expression.

Subtract $2$ from $1$ .

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

Move the negative in front of the fraction.

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

To write $- 9$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Combine $- 9$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 9$ by $3$ .

$f (- \frac{1}{3}) = \frac{- 27 - 1}{3}$

Subtract $1$ from $- 27$ .

$f (- \frac{1}{3}) = \frac{- 28}{3}$

Move the negative in front of the fraction.

$f (- \frac{1}{3}) = - \frac{28}{3}$

The final answer is $- \frac{28}{3}$ .

$- \frac{28}{3}$

Step 4

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

$(- \frac{1}{3}, - \frac{28}{3})$

Step 5

Find the Maximum/Minimum Value f(x)=3x^2+2x-9

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