Precalculus Examples

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

Step 1

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

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

Step 2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y - 1 = \pm \sqrt{16 - x^{2}}$

Step 3

Simplify $\pm \sqrt{16 - x^{2}}$ .

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Step 3.1

Rewrite $16$ as $4^{2}$ .

$y - 1 = \pm \sqrt{4^{2} - x^{2}}$

Step 3.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = x$ .

$y - 1 = \pm \sqrt{(4 + x) (4 - x)}$

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 4.1

First, use the positive value of the $\pm$ to find the first solution.

$y - 1 = \sqrt{(4 + x) (4 - x)}$

Step 4.2

Add $1$ to both sides of the equation.

$y = \sqrt{(4 + x) (4 - x)} + 1$

Step 4.3

Next, use the negative value of the $\pm$ to find the second solution.

$y - 1 = - \sqrt{(4 + x) (4 - x)}$

Step 4.4

Add $1$ to both sides of the equation.

$y = - \sqrt{(4 + x) (4 - x)} + 1$

Step 4.5

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{(4 + x) (4 - x)} + 1$

$y = - \sqrt{(4 + x) (4 - x)} + 1$

$y = \sqrt{(4 + x) (4 - x)} + 1$

$y = - \sqrt{(4 + x) (4 - x)} + 1$

Step 5

To rewrite as a function of $x$ , write the equation so that $y$ is by itself on one side of the equal sign and an expression involving only $x$ is on the other side.

$f (x) = \sqrt{(4 + x) (4 - x)} + 1, - \sqrt{(4 + x) (4 - x)} + 1$