Find the Vertex Form f(x)=4+x^2-2x

$f (x) = 4 + x^{2} - 2 x$

Step 1

Write $f (x) = 4 + x^{2} - 2 x$ as an equation.

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

Step 2

Move $4$ .

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

Step 3

Complete the square for $x^{2} - 2 x + 4$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = - 2$

$c = 4$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

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

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

Factor $2$ out of $- 2$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 1$ by $1$ .

$d = - 1$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 4 - \frac{{(- 2)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

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

$e = 4 - \frac{4}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 4 - \frac{4}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$e = 4 - \frac{4}{4}$

Rewrite the expression.

$e = 4 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = 4 - 1$

Subtract $1$ from $4$ .

$e = 3$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x - 1)}^{2} + 3$ .

${(x - 1)}^{2} + 3$

Step 4

Set $y$ equal to the new right side.

$y = {(x - 1)}^{2} + 3$

Find the Vertex Form f(x)=4+x^2-2x

Related Questions

Support LCMGCF.com by disabling your adblocker.