Find the Standard Form of the Parabola x^2-2x-y-4=0

$x^{2} - 2 x - y - 4 = 0$

Move all terms not containing $x$ to the right side of the equation.

Add $y$ to both sides of the equation.

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

Add $4$ to both sides of the equation.

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

Complete the square.

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

$a = 1$

$b = - 2$

$c = 0 = y + 4$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = y + 4$

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 2)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

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

$e = 0 - \frac{4}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{4}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$e = 0 - \frac{4}{4}$

Rewrite the expression.

$e = 0 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = 0 - 1$

Subtract $1$ from $0$ .

$e = - 1$

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

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

Move all terms not containing $x$ to the right side of the equation.

Add $1$ to both sides of the equation.

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

Add $4$ and $1$ .

${(x - 1)}^{2} = y + 5$

Find the Standard Form of the Parabola x^2-2x-y-4=0

Related Questions

Support LCMGCF.com by disabling your adblocker.