Find the Maximum/Minimum Value y=x^2-4x+1

$y = x^{2} - 4 x + 1$

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

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

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

$x = - \frac{- 4}{2 (1)}$

Remove parentheses.

$x = - \frac{- 4}{2 (1)}$

Simplify $- \frac{- 4}{2 (1)}$ .

Cancel the common factor of $- 4$ and $2$ .

Factor $2$ out of $- 4$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 2$ by $1$ .

$x = - - 2$

Multiply $- 1$ by $- 2$ .

$x = 2$

Evaluate $f (2)$ .

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

$f (2) = {(2)}^{2} - 4 \cdot 2 + 1$

Simplify the result.

Simplify each term.

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

$f (2) = 4 - 4 \cdot 2 + 1$

Multiply $- 4$ by $2$ .

$f (2) = 4 - 8 + 1$

Simplify by adding and subtracting.

Subtract $8$ from $4$ .

$f (2) = - 4 + 1$

Add $- 4$ and $1$ .

$f (2) = - 3$

The final answer is $- 3$ .

$- 3$

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

$(2, - 3)$

Find the Maximum/Minimum Value y=x^2-4x+1

Related Questions

Support LCMGCF.com by disabling your adblocker.