Find the Maximum/Minimum Value y=x^2-6x-4

$y = x^{2} - 6 x - 4$

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 $b$ .

$x = - \frac{- 6}{2 (1)}$

Remove parentheses.

$x = - \frac{- 6}{2 (1)}$

Simplify $- \frac{- 6}{2 (1)}$ .

Cancel the common factor of $- 6$ and $2$ .

Factor $2$ out of $- 6$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

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

Cancel the common factor.

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

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.

$f (3) = {(3)}^{2} - 6 \cdot 3 - 4$

Simplify the result.

Simplify each term.

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

$f (3) = 9 - 6 \cdot 3 - 4$

Multiply $- 6$ by $3$ .

$f (3) = 9 - 18 - 4$

Simplify by subtracting numbers.

Subtract $18$ from $9$ .

$f (3) = - 9 - 4$

Subtract $4$ from $- 9$ .

$f (3) = - 13$

The final answer is $- 13$ .

$- 13$

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

$(3, - 13)$

Find the Maximum/Minimum Value y=x^2-6x-4

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