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

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

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 $b$ .

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

Remove parentheses.

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

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

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

Factor $2$ out of $- 2$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 1$ by $1$ .

$x = - - 1$

Multiply $- 1$ by $- 1$ .

$x = 1$

Evaluate $f (1)$ .

Replace the variable $x$ with $1$ in the expression.

$f (1) = {(1)}^{2} - 2 \cdot 1$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = 1 - 2 \cdot 1$

Multiply $- 2$ by $1$ .

$f (1) = 1 - 2$

Subtract $2$ from $1$ .

$f (1) = - 1$

The final answer is $- 1$ .

$- 1$

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

$(1, - 1)$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.