Find the Standard Form of the Parabola x^2+2x-4y+17=0

$x^{2} + 2 x - 4 y + 17 = 0$

Step 1

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

Add $4 y$ to both sides of the equation.

$x^{2} + 2 x + 17 = 4 y$

Subtract $17$ from both sides of the equation.

$x^{2} + 2 x = 4 y - 17$

Step 2

Complete the square.

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

$a = 1$

$b = 2$

$c = 0 = 4 y - 17$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = 4 y - 17$

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 $2$ .

Cancel the common factor.

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

Rewrite the expression.

$d = 1$

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

${(x + 1)}^{2} - 1 = 4 y - 17$

Step 3

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

Add $1$ to both sides of the equation.

${(x + 1)}^{2} = 4 y - 17 + 1$

Add $- 17$ and $1$ .

${(x + 1)}^{2} = 4 y - 16$

Step 4

Factor $4$ out of $4 y - 16$ .

Factor $4$ out of $4 y$ .

${(x + 1)}^{2} = 4 (y) - 16$

Factor $4$ out of $- 16$ .

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

Factor $4$ out of $4 y + 4 \cdot - 4$ .

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

Find the Standard Form of the Parabola x^2+2x-4y+17=0

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