Find the Maximum/Minimum Value 3x^2-18x+28

$3 x^{2} - 18 x + 28$

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

Find the value of $x = - \frac{b}{2 a}$ .

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

$x = - \frac{- 18}{2 (3)}$

Remove parentheses.

$x = - \frac{- 18}{2 (3)}$

Simplify $- \frac{- 18}{2 (3)}$ .

Cancel the common factor of $- 18$ and $2$ .

Factor $2$ out of $- 18$ .

$x = - \frac{2 \cdot - 9}{2 \cdot 3}$

Cancel the common factors.

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

$x = - \frac{2 \cdot - 9}{2 (3)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 9}{2 \cdot 3}$

Rewrite the expression.

$x = - \frac{- 9}{3}$

Cancel the common factor of $- 9$ and $3$ .

Factor $3$ out of $- 9$ .

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

Cancel the common factors.

Factor $3$ out of $3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 3$ by $1$ .

$x = - - 3$

Multiply $- 1$ by $- 3$ .

$x = 3$

Evaluate $f (3)$ .

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

$f (3) = 3 {(3)}^{2} - 18 \cdot 3 + 28$

Simplify the result.

Simplify each term.

Multiply $3$ by ${(3)}^{2}$ by adding the exponents.

Multiply $3$ by ${(3)}^{2}$ .

Raise $3$ to the power of $1$ .

$f (3) = 3 {(3)}^{2} - 18 \cdot 3 + 28$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (3) = 3^{1 + 2} - 18 \cdot 3 + 28$

Add $1$ and $2$ .

$f (3) = 3^{3} - 18 \cdot 3 + 28$

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

$f (3) = 27 - 18 \cdot 3 + 28$

Multiply $- 18$ by $3$ .

$f (3) = 27 - 54 + 28$

Simplify by adding and subtracting.

Subtract $54$ from $27$ .

$f (3) = - 27 + 28$

Add $- 27$ and $28$ .

$f (3) = 1$

The final answer is $1$ .

$1$

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

$(3, 1)$

Find the Maximum/Minimum Value 3x^2-18x+28

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