Find the Standard Form of the Parabola x^2+10x-y+29=0

$x^{2} + 10 x - y + 29 = 0$

Step 1

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

Add $y$ to both sides of the equation.

$x^{2} + 10 x + 29 = y$

Subtract $29$ from both sides of the equation.

$x^{2} + 10 x = y - 29$

Step 2

Complete the square.

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

$a = 1$

$b = 10$

$c = 0 = y - 29$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = y - 29$

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

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

$d = \frac{10}{2 \cdot 1}$

Cancel the common factor of $10$ and $2$ .

Factor $2$ out of $10$ .

$d = \frac{2 \cdot 5}{2 \cdot 1}$

Cancel the common factors.

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

$d = \frac{2 \cdot 5}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot 5}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{5}{1}$

Divide $5$ by $1$ .

$d = 5$

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

$e = 0 - \frac{10^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $10$ to the power of $2$ .

$e = 0 - \frac{100}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{100}{4}$

Divide $100$ by $4$ .

$e = 0 - 1 \cdot 25$

Multiply $- 1$ by $25$ .

$e = 0 - 25$

Subtract $25$ from $0$ .

$e = - 25$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 5)}^{2} - 25$ .

${(x + 5)}^{2} - 25 = y - 29$

Step 3

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

Add $25$ to both sides of the equation.

${(x + 5)}^{2} = y - 29 + 25$

Add $- 29$ and $25$ .

${(x + 5)}^{2} = y - 4$

Find the Standard Form of the Parabola x^2+10x-y+29=0

Related Questions

Support LCMGCF.com by disabling your adblocker.