Find the Standard Form of the Parabola x=-17y^2+102y-159

$x = - 17 y^{2} + 102 y - 159$

Since $y$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- 17 y^{2} + 102 y - 159 = x$

Add $159$ to both sides of the equation.

$- 17 y^{2} + 102 y = x + 159$

Complete the square.

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

$a = - 17$

$b = 102$

$c = 0 = x + 159$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = x + 159$

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

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

$d = \frac{102}{2 \cdot - 17}$

Simplify the right side.

Cancel the common factor of $102$ and $2$ .

Factor $2$ out of $102$ .

$d = \frac{2 \cdot 51}{2 \cdot - 17}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 17$ .

$d = \frac{2 \cdot 51}{2 (- 17)}$

Cancel the common factor.

$d = \frac{2 \cdot 51}{2 \cdot - 17}$

Rewrite the expression.

$d = \frac{51}{- 17}$

Cancel the common factor of $51$ and $- 17$ .

Factor $17$ out of $51$ .

$d = \frac{17 (3)}{- 17}$

Move the negative one from the denominator of $\frac{3}{- 1}$ .

$d = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$d = - 3$

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

$e = 0 - \frac{102^{2}}{4 \cdot - 17}$

Simplify the right side.

Simplify each term.

Raise $102$ to the power of $2$ .

$e = 0 - \frac{10404}{4 \cdot - 17}$

Multiply $4$ by $- 17$ .

$e = 0 - \frac{10404}{- 68}$

Divide $10404$ by $- 68$ .

$e = 0 - - 153$

Multiply $- 1$ by $- 153$ .

$e = 0 + 153$

Add $0$ and $153$ .

$e = 153$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 17 {(y - 3)}^{2} + 153$ .

$- 17 {(y - 3)}^{2} + 153 = x + 159$

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

Subtract $153$ from both sides of the equation.

$- 17 {(y - 3)}^{2} = x + 159 - 153$

Subtract $153$ from $159$ .

$- 17 {(y - 3)}^{2} = x + 6$

Divide each term in $- 17 {(y - 3)}^{2} = x + 6$ by $- 17$ and simplify.

Divide each term in $- 17 {(y - 3)}^{2} = x + 6$ by $- 17$ .

$\frac{- 17 {(y - 3)}^{2}}{- 17} = \frac{x}{- 17} + \frac{6}{- 17}$

Simplify the left side.

Cancel the common factor of $- 17$ .

Cancel the common factor.

$\frac{- 17 {(y - 3)}^{2}}{- 17} = \frac{x}{- 17} + \frac{6}{- 17}$

Divide ${(y - 3)}^{2}$ by $1$ .

${(y - 3)}^{2} = \frac{x}{- 17} + \frac{6}{- 17}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

${(y - 3)}^{2} = - \frac{x}{17} + \frac{6}{- 17}$

Move the negative in front of the fraction.

${(y - 3)}^{2} = - \frac{x}{17} - \frac{6}{17}$

Factor.

Factor $- 1$ out of $- \frac{x}{17} - \frac{6}{17}$ .

Factor $- 1$ out of $- \frac{x}{17}$ .

${(y - 3)}^{2} = - (\frac{x}{17}) - \frac{6}{17}$

Factor $- 1$ out of $- \frac{6}{17}$ .

${(y - 3)}^{2} = - (\frac{x}{17}) - (\frac{6}{17})$

Factor $- 1$ out of $- (\frac{x}{17}) - (\frac{6}{17})$ .

${(y - 3)}^{2} = - (\frac{x}{17} + \frac{6}{17})$

Combine the numerators over the common denominator.

${(y - 3)}^{2} = - \frac{x + 6}{17}$

Reorder terms.

${(y - 3)}^{2} = - (\frac{1}{17} (x + 6))$

Remove parentheses.

${(y - 3)}^{2} = - \frac{1}{17} (x + 6)$

Find the Standard Form of the Parabola x=-17y^2+102y-159

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