Find the Maximum/Minimum Value f(x)=9x^2-36x+6

$f (x) = 9 x^{2} - 36 x + 6$

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 $b$ .

$x = - \frac{- 36}{2 (9)}$

Remove parentheses.

$x = - \frac{- 36}{2 (9)}$

Simplify $- \frac{- 36}{2 (9)}$ .

Cancel the common factor of $- 36$ and $2$ .

Factor $2$ out of $- 36$ .

$x = - \frac{2 \cdot - 18}{2 \cdot 9}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 9$ .

$x = - \frac{2 \cdot - 18}{2 (9)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 18}{2 \cdot 9}$

Rewrite the expression.

$x = - \frac{- 18}{9}$

Cancel the common factor of $- 18$ and $9$ .

Factor $9$ out of $- 18$ .

$x = - \frac{9 \cdot - 2}{9}$

Cancel the common factors.

Factor $9$ out of $9$ .

$x = - \frac{9 \cdot - 2}{9 (1)}$

Cancel the common factor.

$x = - \frac{9 \cdot - 2}{9 \cdot 1}$

Rewrite the expression.

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

Divide $- 2$ by $1$ .

$x = - - 2$

Multiply $- 1$ by $- 2$ .

$x = 2$

Step 3

Evaluate $f (2)$ .

Replace the variable $x$ with $2$ in the expression.

$f (2) = 9 {(2)}^{2} - 36 \cdot 2 + 6$

Simplify the result.

Simplify each term.

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

$f (2) = 9 \cdot 4 - 36 \cdot 2 + 6$

Multiply $9$ by $4$ .

$f (2) = 36 - 36 \cdot 2 + 6$

Multiply $- 36$ by $2$ .

$f (2) = 36 - 72 + 6$

Simplify by adding and subtracting.

Subtract $72$ from $36$ .

$f (2) = - 36 + 6$

Add $- 36$ and $6$ .

$f (2) = - 30$

The final answer is $- 30$ .

$- 30$

Step 4

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

$(2, - 30)$

Step 5

Find the Maximum/Minimum Value f(x)=9x^2-36x+6

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