Precalculus Examples

$x^{2} - 8 x - y + 17 = 0$

Step 1

Solve for $y$ .

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Step 1.1

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

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Step 1.1.1

Subtract $x^{2}$ from both sides of the equation.

$- 8 x - y + 17 = - x^{2}$

Step 1.1.2

Add $8 x$ to both sides of the equation.

$- y + 17 = - x^{2} + 8 x$

Step 1.1.3

Subtract $17$ from both sides of the equation.

$- y = - x^{2} + 8 x - 17$

Step 1.2

Divide each term in $- y = - x^{2} + 8 x - 17$ by $- 1$ and simplify.

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Step 1.2.1

Divide each term in $- y = - x^{2} + 8 x - 17$ by $- 1$ .

$\frac{- y}{- 1} = \frac{- x^{2}}{- 1} + \frac{8 x}{- 1} + \frac{- 17}{- 1}$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{- x^{2}}{- 1} + \frac{8 x}{- 1} + \frac{- 17}{- 1}$

Step 1.2.2.2

Divide $y$ by $1$ .

$y = \frac{- x^{2}}{- 1} + \frac{8 x}{- 1} + \frac{- 17}{- 1}$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Dividing two negative values results in a positive value.

$y = \frac{x^{2}}{1} + \frac{8 x}{- 1} + \frac{- 17}{- 1}$

Step 1.2.3.1.2

Divide $x^{2}$ by $1$ .

$y = x^{2} + \frac{8 x}{- 1} + \frac{- 17}{- 1}$

Step 1.2.3.1.3

Move the negative one from the denominator of $\frac{8 x}{- 1}$ .

$y = x^{2} - 1 \cdot (8 x) + \frac{- 17}{- 1}$

Step 1.2.3.1.4

Rewrite $- 1 \cdot (8 x)$ as $- (8 x)$ .

$y = x^{2} - (8 x) + \frac{- 17}{- 1}$

Step 1.2.3.1.5

Multiply $8$ by $- 1$ .

$y = x^{2} - 8 x + \frac{- 17}{- 1}$

Step 1.2.3.1.6

Divide $- 17$ by $- 1$ .

$y = x^{2} - 8 x + 17$

Step 2

To write a polynomial in standard form, simplify and then arrange the terms in descending order.

$y = a x^{2} + b x + c$

Step 3

The standard form is $x^{2} - 8 x + 17$ .

$y = x^{2} - 8 x + 17$

Step 4