Precalculus Examples

$f (x) = x^{2} - 6 x + k$

Step 1

Find the standard form of the hyperbola.

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Step 1.1

Move all terms containing variables to the left side of the equation.

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Step 1.1.1

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

$y - x^{2} = - 6 x + k$

Step 1.1.2

Add $6 x$ to both sides of the equation.

$y - x^{2} + 6 x = k$

Step 1.1.3

Subtract $k$ from both sides of the equation.

$y - x^{2} + 6 x - k = 0$

Step 1.1.4

Move $6 x$ .

$y - x^{2} - k + 6 x = 0$

Step 1.1.5

Move $y$ .

$- x^{2} - k + y + 6 x = 0$

Step 1.2

Complete the square for $- x^{2} + 6 x$ .

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Step 1.2.1

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

$a = - 1$

$b = 6$

$c = 0$

Step 1.2.2

Consider the vertex form of a parabola.

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

Step 1.2.3

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

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Step 1.2.3.1

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

$d = \frac{6}{2 \cdot - 1}$

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

Cancel the common factor of $6$ and $2$ .

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Step 1.2.3.2.1.1

Factor $2$ out of $6$ .

$d = \frac{2 \cdot 3}{2 \cdot - 1}$

Step 1.2.3.2.1.2

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

$d = - 1 \cdot 3$

Step 1.2.3.2.2

Multiply $- 1$ by $3$ .

$d = - 3$

Step 1.2.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 1.2.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{6^{2}}{4 \cdot - 1}$

Step 1.2.4.2

Simplify the right side.

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Step 1.2.4.2.1

Simplify each term.

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Step 1.2.4.2.1.1

Raise $6$ to the power of $2$ .

$e = 0 - \frac{36}{4 \cdot - 1}$

Step 1.2.4.2.1.2

Multiply $4$ by $- 1$ .

$e = 0 - \frac{36}{- 4}$

Step 1.2.4.2.1.3

Divide $36$ by $- 4$ .

$e = 0 - - 9$

Step 1.2.4.2.1.4

Multiply $- 1$ by $- 9$ .

$e = 0 + 9$

Step 1.2.4.2.2

Add $0$ and $9$ .

$e = 9$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x - 3)}^{2} + 9$ .

$- {(x - 3)}^{2} + 9$

Step 1.3

Substitute $- {(x - 3)}^{2} + 9$ for $- x^{2} + 6 x$ in the equation $- x^{2} - k + y + 6 x = 0$ .

$- {(x - 3)}^{2} + 9 - k + y = 0$

Step 1.4

Move $9$ to the right side of the equation by adding $9$ to both sides.

$- {(x - 3)}^{2} - k + y = 0 - 9$

Step 1.5

Subtract $9$ from $0$ .

$- {(x - 3)}^{2} - k + y = - 9$

Step 2

This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.

$\frac{{(y - k)}^{2}}{a^{2}} - \frac{{(x - h)}^{2}}{b^{2}} = 1$

Step 3

Match the values in this hyperbola to those of the standard form. The variable $h$ represents the x-offset from the origin, $k$ represents the y-offset from origin, $a$ .

$a = 1$

$b = 1$

$k = 0$

$h = 3$

Step 4

The center of a hyperbola follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(3, 0)$

Step 5

Find $c$ , the distance from the center to a focus.

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Step 5.1

Find the distance from the center to a focus of the hyperbola by using the following formula.

$\sqrt{a^{2} + b^{2}}$

Step 5.2

Substitute the values of $a$ and $b$ in the formula.

$\sqrt{{(1)}^{2} + {(1)}^{2}}$

Step 5.3

Simplify.

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Step 5.3.1

One to any power is one.

$\sqrt{1 + {(1)}^{2}}$

Step 5.3.2

One to any power is one.

$\sqrt{1 + 1}$

Step 5.3.3

Add $1$ and $1$ .

$\sqrt{2}$

Step 6

Find the vertices.

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Step 6.1

The first vertex of a hyperbola can be found by adding $a$ to $k$ .

$(h, k + a)$

Step 6.2

Substitute the known values of $h$ , $a$ , and $k$ into the formula and simplify.

$(3, 1)$

Step 6.3

The second vertex of a hyperbola can be found by subtracting $a$ from $k$ .

$(h, k - a)$

Step 6.4

Substitute the known values of $h$ , $a$ , and $k$ into the formula and simplify.

$(3, - 1)$

Step 6.5

The vertices of a hyperbola follow the form of $(h, k \pm a)$ . Hyperbolas have two vertices.

$(3, 1), (3, - 1)$

Step 7

Find the foci.

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Step 7.1

The first focus of a hyperbola can be found by adding $c$ to $k$ .

$(h, k + c)$

Step 7.2

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(3, \sqrt{2})$

Step 7.3

The second focus of a hyperbola can be found by subtracting $c$ from $k$ .

$(h, k - c)$

Step 7.4

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(3, - \sqrt{2})$

Step 7.5

The foci of a hyperbola follow the form of $(h, k \pm \sqrt{a^{2} + b^{2}})$ . Hyperbolas have two foci.

$(3, \sqrt{2}), (3, - \sqrt{2})$

Step 8

Find the focal parameter.

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Step 8.1

Find the value of the focal parameter of the hyperbola by using the following formula.

$\frac{b^{2}}{\sqrt{a^{2} + b^{2}}}$

Step 8.2

Substitute the values of $b$ and $\sqrt{a^{2} + b^{2}}$ in the formula.

$\frac{1^{2}}{\sqrt{2}}$

Step 8.3

Simplify.

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Step 8.3.1

One to any power is one.

$\frac{1}{\sqrt{2}}$

Step 8.3.2

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Step 8.3.3

Combine and simplify the denominator.

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Step 8.3.3.1

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\frac{\sqrt{2}}{\sqrt{2} \sqrt{2}}$

Step 8.3.3.2

Raise $\sqrt{2}$ to the power of $1$ .

$\frac{\sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

Step 8.3.3.3

Raise $\sqrt{2}$ to the power of $1$ .

$\frac{\sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

Step 8.3.3.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Step 8.3.3.5

Add $1$ and $1$ .

$\frac{\sqrt{2}}{{\sqrt{2}}^{2}}$

Step 8.3.3.6

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Step 8.3.3.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$\frac{\sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

Step 8.3.3.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{\sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Step 8.3.3.6.3

Combine $\frac{1}{2}$ and $2$ .

$\frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Step 8.3.3.6.4

Cancel the common factor of $2$ .

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Step 8.3.3.6.4.1

Cancel the common factor.

$\frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Step 8.3.3.6.4.2

Rewrite the expression.

$\frac{\sqrt{2}}{2^{1}}$

Step 8.3.3.6.5

Evaluate the exponent.

$\frac{\sqrt{2}}{2}$

Step 9

The asymptotes follow the form $y = \pm \frac{a (x - h)}{b} + k$ because this hyperbola opens up and down.

$y = \pm 1 \cdot (x - (3)) + 0$

Step 10

Simplify to find the first asymptote.

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Step 10.1

Remove parentheses.

$y = 1 \cdot (x - (3)) + 0$

Step 10.2

Simplify $1 \cdot (x - (3)) + 0$ .

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Step 10.2.1

Add $1 \cdot (x - (3))$ and $0$ .

$y = 1 \cdot (x - (3))$

Step 10.2.2

Multiply $x - (3)$ by $1$ .

$y = x - (3)$

Step 10.2.3

Multiply $- 1$ by $3$ .

$y = x - 3$

Step 11

Simplify to find the second asymptote.

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Step 11.1

Remove parentheses.

$y = - 1 \cdot (x - (3)) + 0$

Step 11.2

Simplify $- 1 \cdot (x - (3)) + 0$ .

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Step 11.2.1

Simplify the expression.

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Step 11.2.1.1

Add $- 1 \cdot (x - (3))$ and $0$ .

$y = - 1 \cdot (x - (3))$

Step 11.2.1.2

Multiply $- 1$ by $3$ .

$y = - 1 \cdot (x - 3)$

Step 11.2.2

Apply the distributive property.

$y = - 1 x - 1 \cdot - 3$

Step 11.2.3

Simplify the expression.

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Step 11.2.3.1

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

$y = - x - 1 \cdot - 3$

Step 11.2.3.2

Multiply $- 1$ by $- 3$ .

$y = - x + 3$

Step 12

This hyperbola has two asymptotes.

$y = x - 3, y = - x + 3$

Step 13

These values represent the important values for graphing and analyzing a hyperbola.

Center: $(3, 0)$

Vertices: $(3, 1), (3, - 1)$

Foci: $(3, \sqrt{2}), (3, - \sqrt{2})$

Eccentricity: $(3, \sqrt{2}), (3, - \sqrt{2})$

Focal Parameter: $\frac{\sqrt{2}}{2}$

Asymptotes: $y = x - 3$ , $y = - x + 3$

Step 14