Find the Maximum/Minimum Value 3x^2+30x+71

$3 x^{2} + 30 x + 71$

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 $b$ .

$x = - \frac{30}{2 (3)}$

Remove parentheses.

$x = - \frac{30}{2 (3)}$

Simplify $- \frac{30}{2 (3)}$ .

Cancel the common factor of $30$ and $2$ .

Factor $2$ out of $30$ .

$x = - \frac{2 \cdot 15}{2 \cdot 3}$

Cancel the common factors.

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

$x = - \frac{2 \cdot 15}{2 (3)}$

Cancel the common factor.

$x = - \frac{2 \cdot 15}{2 \cdot 3}$

Rewrite the expression.

$x = - \frac{15}{3}$

Cancel the common factor of $15$ and $3$ .

Factor $3$ out of $15$ .

$x = - \frac{3 \cdot 5}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

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

Cancel the common factor.

$x = - \frac{3 \cdot 5}{3 \cdot 1}$

Rewrite the expression.

$x = - \frac{5}{1}$

Divide $5$ by $1$ .

$x = - 1 \cdot 5$

Multiply $- 1$ by $5$ .

$x = - 5$

Step 3

Evaluate $f (- 5)$ .

Replace the variable $x$ with $- 5$ in the expression.

$f (- 5) = 3 {(- 5)}^{2} + 30 (- 5) + 71$

Simplify the result.

Simplify each term.

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

$f (- 5) = 3 \cdot 25 + 30 (- 5) + 71$

Multiply $3$ by $25$ .

$f (- 5) = 75 + 30 (- 5) + 71$

Multiply $30$ by $- 5$ .

$f (- 5) = 75 - 150 + 71$

Simplify by adding and subtracting.

Subtract $150$ from $75$ .

$f (- 5) = - 75 + 71$

Add $- 75$ and $71$ .

$f (- 5) = - 4$

The final answer is $- 4$ .

$- 4$

Step 4

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

$(- 5, - 4)$

Step 5

Find the Maximum/Minimum Value 3x^2+30x+71

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