Algebra II - Hyperbolic Functions

What is the value of $-\tanh(0)$ ?

$\text{Does not exist}$

$\frac{1}{2}$

Explanation

The hyperbolic tangent will need to be rewritten in terms of hyperbolic sine and cosine.

According to the properties:

$\sinh(0)=0$

$\cosh(0)=1$

Therefore:

$-\tanh(0)= -\frac{\sinh(0)}{\cosh(0)}=-\frac{\frac{e^0-e^0}{2}}{\frac{e^0+e^0}{2}}=-\frac{e^0-e^0}{e^0+e^0}=-\frac{0}{1} = 0$

$\small f(x)=\tfrac{1}{x+4}-3$

Which of the following is the correct expression for a hyperbola that is shifted $\small 5$ $\small units$ units up and $\small 2$ $\small units$ to the right of $\small f(x)$ ?

$\small \small \small g(x)=\tfrac{1}{x+2}+2$

$\small \small \small \small g(x)=\tfrac{1}{x+9}-1$

$\small \small \small \small g(x)=\tfrac{1}{x+2}+5$

$\small \small \small \small g(x)=\tfrac{1}{x-2}+5$

$\small \small \small \small g(x)=\tfrac{1}{x+6}-8$

Explanation

The parent function of a hyperbola is represented by the function $\small \small \small \small g(x)=\tfrac{1}{x-h}+k$ where $\small (h,k)$ is the center of the hyperbola. To shift the original function up by $\small 5$ $\small units$ simply add $\small -3+5$ . To shift it to the right $\small 2$ $\small units$ take away $\small 4-2$ .

What is the shape of the graph depicted by the equation:

$\small \frac{y^2}{9}-\frac{x^2}{4}=1$

Hyperbola

Parabola

Circle

Oval

Explanation

The standard equation of a hyperbola is:

$\small \frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

Express the following hyperbolic function in standard form:

$\frac{(y-7)^2}{4}-\frac{(x+6)^2}{4}=1$

$\frac{(y-7)^2}{13}+\frac{(x+6)^2}{13}=1$

$\frac{(x-7)^2}{52}-\frac{(y+6)^2}{52}=1$

$\frac{(x-7)^2}{4}+\frac{(y+6)^2}{4}=1$

$\frac{2(y-7)^2}{13}-\frac{2(x+6)^2}{13}=1$

Explanation

In order to express the given hyperbolic function in standard form, we must write it in one of the following two ways:

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

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

From our formulas for the standard form of a hyperbolic equation above, we can see that the term on the right side of the equation is always 1, so we must divide both sides of the given equation by 52, which gives us:

$\frac{13(y-7)^2}{52}-\frac{13(x+6)^2}{52}=1$

Simplifying, we obtain our final answer in standard form:

$\frac{(y-7)^2}{4}-\frac{(x+6)^2}{4}=1$

Given the hyperbola , what is the value of the center?

Explanation

In order to determine the center, we will first need to rewrite this equation in standard form.

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

Isolate 41 on the right side. Subtract and add on both sides.

The equation becomes:

Group the x and y terms. Be careful of the negative signs.

Pull out a common factor of 4 on the second parentheses.

Complete the square twice. Divide the second term of each parentheses by two and square the quantity. Add the terms on both sides.

$(y^2+4y+(\frac{4}{2})^2)-4(x^2-6x+(\frac{-6}{3})^2)= 41+(\frac{4}{2})^2+(-4)(\frac{-6}{3})^2$

This equation becomes:

Factorize the left side and simplify the right.

Divide both sides by nine.

$\frac{(y+2)^2}{9}-\frac{4}{9}(x-3)^2 = \frac{9}{9}$

$\frac{(y+2)^2}{9}-\frac{4(x-3)^2 }{9}=1$

The equation is now in the standard form of a hyperbola.

The center is at:

The answer is:

Find the foci of the hyperbola: $-\frac{x^2}{36}+\frac{y^2}{16} =1$

$(0,\pm 2\sqrt{13})$

$(0,\pm4)$

$(\pm6,0)$

$(\pm 2\sqrt{13},0)$

$(\pm 4\sqrt{13},0)$

Explanation

Write the standard forms for a hyperbola.

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

OR:

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

The standard form is given in the second case, which will have different parameters compared to the first form.

Center:

Foci: $(h,k\pm c)$ , where

$-\frac{x^2}{36}+\frac{y^2}{16} =1$

Identify the coefficients and substitute to find the value of .

$a^2 = 36, a=\pm6$

$b^2 =16, b=\pm 4$

$c^2 =a^2+b^2\rightarrow c^2 =36+16$

$c^2=52, c=\pm \sqrt{52} =\pm \sqrt{4}\cdot \sqrt{13}= \pm2\sqrt{13}$

$(h,k\pm c) = (0,0 \pm 2\sqrt{13}) = (0,\pm 2\sqrt{13})$

The answer is: $(0,\pm 2\sqrt{13})$

Which of the following equations represents a vertical hyperbola with a center of and asymptotes at $y=\pm, \frac{4}{3}, x$ ?

$\frac{(y-3)^{2}}{16}-\frac{(x+4)^{2}}{9}=1$

$\frac{(x+4)^{2}}{9}-\frac{(y-3)^{2}}{16}=1$

$\frac{(y+3)^{2}}{16}-\frac{(x-4)^{2}}{9}=1$

$\frac{(y-3)^{2}}{9}-\frac{(x+4)^{2}}{16}=1$

Explanation

First, we need to become familiar with the standard form of a hyperbolic equation:

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

The center is always at . This means that for this problem, the numerators of each term will have to contain $(x+4)^{2}$ and $(y-3)^{2}$ .

To determine if a hyperbola opens vertically or horizontally, look at the sign of each variable. A vertical parabola has a positive $y^{2}$ term; a horizontal parabola has a positive $x^{2}$ term. In this case, we need a vertical parabola, so the $y^{2}$ term will have to be positive.

(NOTE: If both terms are the same sign, you have an ellipse, not a parabola.)

The asymptotes of a parabola are always found by the equation $y=\pm, \frac{b}{a}, x$ , where is found in the denominator of the $y^{2}$ term and is found in the denominator of the $x^{2}$ term. Since our asymptotes are $y=\pm, \frac{4}{3}, x$ , we know that must be 4 and must be 3. That means that the number underneath the $y^{2}$ term has to be 16, and the number underneath the $x^{2}$ term has to be 9.

In which direction does the graph of the hyperbola open?

vertical

horizontal

Explanation

The equation for a horizontal hyperbola is . The equation for a vertical hyperbola is . The x-term appears first, so the given equation represents a horizontal hyperbola.

Which of the following shapes does the graph of the equation take?

Circle

Ellipse

Parabola

Hyperbola

Explanation

The equation for a horizontal hyperbola is . The equation for a vertical hyperbola is . Both include x- and y-terms combined using subtraction. The equation for an ellipse is . Because the given equation connects the x- and y-terms using addition rather than subtraction, it represents an ellipse rather than a hyperbola. If the equation took the form , it would represent a circle. If the equation took the form , it would represent a parabola.

Which of the following shapes does the graph of the equation take?

Circle

Ellipse

Parabola

Hyperbola

Explanation

The equation for a horizontal hyperbola is . The equation for a vertical hyperbola is . Both include x- and y-terms combined using subtraction. If the equation took the form (using addition rather than subtraction to combine the x- and y-terms), it would represent an ellipse. If the equation took the form , it would represent a circle. If the equation took the form , it would represent a parabola.

Hyperbolic Functions

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