Find the Maximum/Minimum Value y=3x^2-12x+10

$y = 3 x^{2} - 12 x + 10$

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 $b$ .

$x = - \frac{- 12}{2 (3)}$

Remove parentheses.

$x = - \frac{- 12}{2 (3)}$

Simplify $- \frac{- 12}{2 (3)}$ .

Cancel the common factor of $- 12$ and $2$ .

Factor $2$ out of $- 12$ .

$x = - \frac{2 \cdot - 6}{2 \cdot 3}$

Cancel the common factors.

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

$x = - \frac{2 \cdot - 6}{2 (3)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 6}{2 \cdot 3}$

Rewrite the expression.

$x = - \frac{- 6}{3}$

Cancel the common factor of $- 6$ and $3$ .

Factor $3$ out of $- 6$ .

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

Cancel the common factors.

Factor $3$ out of $3$ .

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

Cancel the common factor.

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

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.

$f (2) = 3 {(2)}^{2} - 12 \cdot 2 + 10$

Simplify the result.

Simplify each term.

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

$f (2) = 3 \cdot 4 - 12 \cdot 2 + 10$

Multiply $3$ by $4$ .

$f (2) = 12 - 12 \cdot 2 + 10$

Multiply $- 12$ by $2$ .

$f (2) = 12 - 24 + 10$

Simplify by adding and subtracting.

Subtract $24$ from $12$ .

$f (2) = - 12 + 10$

Add $- 12$ and $10$ .

$f (2) = - 2$

The final answer is $- 2$ .

$- 2$

Step 4

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

$(2, - 2)$

Step 5

Find the Maximum/Minimum Value y=3x^2-12x+10

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