Precalculus Examples

$\frac{{(y + 5)}^{2}}{4} - \frac{{(x - 1)}^{2}}{16} = 1$

Step 1

Add $\frac{{(x - 1)}^{2}}{16}$ to both sides of the equation.

$\frac{{(y + 5)}^{2}}{4} = 1 + \frac{{(x - 1)}^{2}}{16}$

Step 2

Simplify $1 + \frac{{(x - 1)}^{2}}{16}$ .

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Step 2.1

Combine into one fraction.

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Step 2.1.1

Write $1$ as a fraction with a common denominator.

$\frac{{(y + 5)}^{2}}{4} = \frac{16}{16} + \frac{{(x - 1)}^{2}}{16}$

Step 2.1.2

Combine the numerators over the common denominator.

$\frac{{(y + 5)}^{2}}{4} = \frac{16 + {(x - 1)}^{2}}{16}$

Step 2.2

Simplify the numerator.

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Step 2.2.1

Rewrite ${(x - 1)}^{2}$ as $(x - 1) (x - 1)$ .

$\frac{{(y + 5)}^{2}}{4} = \frac{16 + (x - 1) (x - 1)}{16}$

Step 2.2.2

Expand $(x - 1) (x - 1)$ using the FOIL Method.

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Step 2.2.2.1

Apply the distributive property.

$\frac{{(y + 5)}^{2}}{4} = \frac{16 + x (x - 1) - 1 (x - 1)}{16}$

Step 2.2.2.2

Apply the distributive property.

$\frac{{(y + 5)}^{2}}{4} = \frac{16 + x \cdot x + x \cdot - 1 - 1 (x - 1)}{16}$

Step 2.2.2.3

Apply the distributive property.

$\frac{{(y + 5)}^{2}}{4} = \frac{16 + x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1}{16}$

Step 2.2.3

Simplify and combine like terms.

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Step 2.2.3.1

Simplify each term.

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Step 2.2.3.1.1

Multiply $x$ by $x$ .

$\frac{{(y + 5)}^{2}}{4} = \frac{16 + x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1}{16}$

Step 2.2.3.1.2

Move $- 1$ to the left of $x$ .

$\frac{{(y + 5)}^{2}}{4} = \frac{16 + x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1}{16}$

Step 2.2.3.1.3

Rewrite $- 1 x$ as $- x$ .

$\frac{{(y + 5)}^{2}}{4} = \frac{16 + x^{2} - x - 1 x - 1 \cdot - 1}{16}$

Step 2.2.3.1.4

Rewrite $- 1 x$ as $- x$ .

$\frac{{(y + 5)}^{2}}{4} = \frac{16 + x^{2} - x - x - 1 \cdot - 1}{16}$

Step 2.2.3.1.5

Multiply $- 1$ by $- 1$ .

$\frac{{(y + 5)}^{2}}{4} = \frac{16 + x^{2} - x - x + 1}{16}$

Step 2.2.3.2

Subtract $x$ from $- x$ .

$\frac{{(y + 5)}^{2}}{4} = \frac{16 + x^{2} - 2 x + 1}{16}$

Step 2.2.4

Add $16$ and $1$ .

$\frac{{(y + 5)}^{2}}{4} = \frac{x^{2} - 2 x + 17}{16}$

Step 3

Multiply both sides of the equation by $4$ .

$4 \frac{{(y + 5)}^{2}}{4} = 4 \frac{x^{2} - 2 x + 17}{16}$

Step 4

Simplify both sides of the equation.

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Step 4.1

Simplify the left side.

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Step 4.1.1

Cancel the common factor of $4$ .

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Step 4.1.1.1

Cancel the common factor.

$4 \frac{{(y + 5)}^{2}}{4} = 4 \frac{x^{2} - 2 x + 17}{16}$

Step 4.1.1.2

Rewrite the expression.

${(y + 5)}^{2} = 4 \frac{x^{2} - 2 x + 17}{16}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Cancel the common factor of $4$ .

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Step 4.2.1.1

Factor $4$ out of $16$ .

${(y + 5)}^{2} = 4 \frac{x^{2} - 2 x + 17}{4 (4)}$

Step 4.2.1.2

Cancel the common factor.

${(y + 5)}^{2} = 4 \frac{x^{2} - 2 x + 17}{4 \cdot 4}$

Step 4.2.1.3

Rewrite the expression.

${(y + 5)}^{2} = \frac{x^{2} - 2 x + 17}{4}$

Step 5

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

$y + 5 = \pm \sqrt{\frac{x^{2} - 2 x + 17}{4}}$

Step 6

Simplify $\pm \sqrt{\frac{x^{2} - 2 x + 17}{4}}$ .

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Step 6.1

Rewrite $\frac{x^{2} - 2 x + 17}{4}$ as ${(\frac{1}{2})}^{2} (x^{2} - 2 x + 17)$ .

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Step 6.1.1

Factor the perfect power $1^{2}$ out of $x^{2} - 2 x + 17$ .

$y + 5 = \pm \sqrt{\frac{1^{2} (x^{2} - 2 x + 17)}{4}}$

Step 6.1.2

Factor the perfect power $2^{2}$ out of $4$ .

$y + 5 = \pm \sqrt{\frac{1^{2} (x^{2} - 2 x + 17)}{2^{2} \cdot 1}}$

Step 6.1.3

Rearrange the fraction $\frac{1^{2} (x^{2} - 2 x + 17)}{2^{2} \cdot 1}$ .

$y + 5 = \pm \sqrt{{(\frac{1}{2})}^{2} (x^{2} - 2 x + 17)}$

Step 6.2

Pull terms out from under the radical.

$y + 5 = \pm \frac{1}{2} \sqrt{x^{2} - 2 x + 17}$

Step 6.3

Combine $\frac{1}{2}$ and $\sqrt{x^{2} - 2 x + 17}$ .

$y + 5 = \pm \frac{\sqrt{x^{2} - 2 x + 17}}{2}$

Step 7

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

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Step 7.1

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

$y + 5 = \frac{\sqrt{x^{2} - 2 x + 17}}{2}$

Step 7.2

Subtract $5$ from both sides of the equation.

$y = \frac{\sqrt{x^{2} - 2 x + 17}}{2} - 5$

Step 7.3

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

$y + 5 = - \frac{\sqrt{x^{2} - 2 x + 17}}{2}$

Step 7.4

Subtract $5$ from both sides of the equation.

$y = - \frac{\sqrt{x^{2} - 2 x + 17}}{2} - 5$

Step 7.5

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

$y = \frac{\sqrt{x^{2} - 2 x + 17}}{2} - 5$

$y = - \frac{\sqrt{x^{2} - 2 x + 17}}{2} - 5$

$y = \frac{\sqrt{x^{2} - 2 x + 17}}{2} - 5$

$y = - \frac{\sqrt{x^{2} - 2 x + 17}}{2} - 5$

Step 8

Set the radicand in $\sqrt{x^{2} - 2 x + 17}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} - 2 x + 17 \geq 0$

Step 9

Solve for $x$ .

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Step 9.1

Convert the inequality to an equation.

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

Step 9.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 9.3

Substitute the values $a = 1$ , $b = - 2$ , and $c = 17$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot 17)}}{2 \cdot 1}$

Step 9.4

Simplify.

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Step 9.4.1

Simplify the numerator.

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Step 9.4.1.1

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 17}}{2 \cdot 1}$

Step 9.4.1.2

Multiply $- 4 \cdot 1 \cdot 17$ .

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Step 9.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 17}}{2 \cdot 1}$

Step 9.4.1.2.2

Multiply $- 4$ by $17$ .

$x = \frac{2 \pm \sqrt{4 - 68}}{2 \cdot 1}$

Step 9.4.1.3

Subtract $68$ from $4$ .

$x = \frac{2 \pm \sqrt{- 64}}{2 \cdot 1}$

Step 9.4.1.4

Rewrite $- 64$ as $- 1 (64)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 64}}{2 \cdot 1}$

Step 9.4.1.5

Rewrite $\sqrt{- 1 (64)}$ as $\sqrt{- 1} \cdot \sqrt{64}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{64}}{2 \cdot 1}$

Step 9.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{64}}{2 \cdot 1}$

Step 9.4.1.7

Rewrite $64$ as $8^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{8^{2}}}{2 \cdot 1}$

Step 9.4.1.8

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{2 \pm i \cdot 8}{2 \cdot 1}$

Step 9.4.1.9

Move $8$ to the left of $i$ .

$x = \frac{2 \pm 8 i}{2 \cdot 1}$

Step 9.4.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 8 i}{2}$

Step 9.4.3

Simplify $\frac{2 \pm 8 i}{2}$ .

$x = 1 \pm 4 i$

Step 9.5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 9.5.1

Simplify the numerator.

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Step 9.5.1.1

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 17}}{2 \cdot 1}$

Step 9.5.1.2

Multiply $- 4 \cdot 1 \cdot 17$ .

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Step 9.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 17}}{2 \cdot 1}$

Step 9.5.1.2.2

Multiply $- 4$ by $17$ .

$x = \frac{2 \pm \sqrt{4 - 68}}{2 \cdot 1}$

Step 9.5.1.3

Subtract $68$ from $4$ .

$x = \frac{2 \pm \sqrt{- 64}}{2 \cdot 1}$

Step 9.5.1.4

Rewrite $- 64$ as $- 1 (64)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 64}}{2 \cdot 1}$

Step 9.5.1.5

Rewrite $\sqrt{- 1 (64)}$ as $\sqrt{- 1} \cdot \sqrt{64}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{64}}{2 \cdot 1}$

Step 9.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{64}}{2 \cdot 1}$

Step 9.5.1.7

Rewrite $64$ as $8^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{8^{2}}}{2 \cdot 1}$

Step 9.5.1.8

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{2 \pm i \cdot 8}{2 \cdot 1}$

Step 9.5.1.9

Move $8$ to the left of $i$ .

$x = \frac{2 \pm 8 i}{2 \cdot 1}$

Step 9.5.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 8 i}{2}$

Step 9.5.3

Simplify $\frac{2 \pm 8 i}{2}$ .

$x = 1 \pm 4 i$

Step 9.5.4

Change the $\pm$ to $+$ .

$x = 1 + 4 i$

Step 9.6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 9.6.1

Simplify the numerator.

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Step 9.6.1.1

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 17}}{2 \cdot 1}$

Step 9.6.1.2

Multiply $- 4 \cdot 1 \cdot 17$ .

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Step 9.6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 17}}{2 \cdot 1}$

Step 9.6.1.2.2

Multiply $- 4$ by $17$ .

$x = \frac{2 \pm \sqrt{4 - 68}}{2 \cdot 1}$

Step 9.6.1.3

Subtract $68$ from $4$ .

$x = \frac{2 \pm \sqrt{- 64}}{2 \cdot 1}$

Step 9.6.1.4

Rewrite $- 64$ as $- 1 (64)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 64}}{2 \cdot 1}$

Step 9.6.1.5

Rewrite $\sqrt{- 1 (64)}$ as $\sqrt{- 1} \cdot \sqrt{64}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{64}}{2 \cdot 1}$

Step 9.6.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{64}}{2 \cdot 1}$

Step 9.6.1.7

Rewrite $64$ as $8^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{8^{2}}}{2 \cdot 1}$

Step 9.6.1.8

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{2 \pm i \cdot 8}{2 \cdot 1}$

Step 9.6.1.9

Move $8$ to the left of $i$ .

$x = \frac{2 \pm 8 i}{2 \cdot 1}$

Step 9.6.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 8 i}{2}$

Step 9.6.3

Simplify $\frac{2 \pm 8 i}{2}$ .

$x = 1 \pm 4 i$

Step 9.6.4

Change the $\pm$ to $-$ .

$x = 1 - 4 i$

Step 9.7

Identify the leading coefficient.

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Step 9.7.1

The leading term in a polynomial is the term with the highest degree.

$x^{2}$

Step 9.7.2

The leading coefficient in a polynomial is the coefficient of the leading term.

$1$

Step 9.8

Since there are no real x-intercepts and the leading coefficient is positive, the parabola opens up and $x^{2} - 2 x + 17$ is always greater than $0$ .

All real numbers

Step 10

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 11

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, - 7] \cup [- 3, \infty)$

Set-Builder Notation:

${y | y \leq - 7, y \geq - 3}$

Step 12

Determine the domain and range.

Domain: $(- \infty, \infty), {x | x \in ℝ}$

Range: $(- \infty, - 7] \cup [- 3, \infty), {y | y \leq - 7, y \geq - 3}$

Step 13