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Algebra II : Functions and Graphs

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Example Questions

Example Question #42 : Circle Functions

Determine the radius of the circle given by the following function:

x^2-4x+y^2-2x=11

Possible Answers:

Correct answer:

Explanation:

To rewrite the given function as the equation of a circle in standard form, we must complete the square for x and y. This method requires us to use the following general form:

(a+b)^2=a^2+2ab+b^2

To start, we can complete the square for the x terms. We must halve the coefficient of x, square it, and add it to the first two terms:

(-2)^2=4

x^2-4x+4

Now, we can rewrite this as a perfect square, but because we added 4, we must subtract 4 as to not change the original function:

(x-2)^2-4

We do the same procedure for the y terms:

(-1)^2=1

y^2-2x+1

(y-1)^2-1

Rewriting our function, we get

(x-2)^2-4+(y-1)^2-1=11

Moving the constants to the right side, we get the function of a circle in standard form:

(x-2)^2+(y-1)^2=16

Comparing to

(x-h)^2+(y-k)^2=r^2

we see that the radius of the circle is

$\sqrt{16}=4$

Notice that the radius is a distance and can therefore never be negative.

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Example Question #491 : Functions And Graphs

$Identify\ the\ center\ and\ radius\ for\ (x+4)^2+(y-2)^2=49$

Possible Answers:

Center: (4,-2); Radius= 49

Center: (-4,2); Radius= 7

Center: (4,-2); Radius= 7

Center: (-4,2); Radius= 49

Correct answer:

Center: (-4,2); Radius= 7

Explanation:

When identifying the center of a circle, take the opposite sign of each value connected to x and y.

$(x+4)^2+(y-2)^2=49\ compare\ with\ the\ general\ equation\ of\ a\ circle: (x-h)^2+(y-k)^2=r^2$

$Changing\ signs\ for\ each\ gives\ us\ a\ center\ of: (-4,2)$

$To\ find\ the\ radius\ we\ set\ r^2=49,$

$take\ square\ root\ of\ both\ sides\ to\ get\ r=7.$

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Example Question #91 : Quadratic Functions

What is the center and radius of the following equation, respectively?

(x-6)^2+(y+7)^2 = 2

Possible Answers:

(6,-7);4

(-6,7); 2

$(-6,7); \sqrt2$

(6,-7); 2

$(6,-7); \sqrt2$

Correct answer:

$(6,-7); \sqrt2$

Explanation:

The equation given represents a circle.

(x-h)^2+(y-k)^2=r^2

(h,k) represents the center, and is the radius.

h=6 , k = -7

The center is at: (6,-7)

Set up an equation to solve the radius.

r^2=2

The radius is: $r=\sqrt{2}$

The answer is: $(6,-7); \sqrt2$

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Example Question #492 : Functions And Graphs

Which of the following represents the formula of a circle with a radius of centered at (-1,2) ?

Possible Answers:

(x+1)^2+(y-2)^2=4

$(x-1)^2+(y+2)^2=2\sqrt2$

$(x-1)^2+(y+2)^2=\sqrt2$

$(x+1)^2+(y-2)^2=\sqrt2$

(x-1)^2+(y+2)^2=4

Correct answer:

(x+1)^2+(y-2)^2=4

Explanation:

Write the standard form for a circle.

(x-h)^2+(y-k)^2=r^2

The circle is centered at: (h,k)

The radius is: r=2

Substitute all the known values into the formula.

(x-(-1))^2+(y-2)^2=2^2

Simplify this equation.

The answer is: (x+1)^2+(y-2)^2=4

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Example Question #1 : Hyperbolic Functions

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

Possible Answers:

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

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

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

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

Correct answer:

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

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 (h,k) . 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.

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Example Question #1 : Hyperbolic Functions

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

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

Possible Answers:

Parabola

Hyperbola

Circle

Oval

Correct answer:

Hyperbola

Explanation:

The standard equation of a hyperbola is:

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

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Example Question #3 : Hyperbolic Functions

Express the following hyperbolic function in standard form:

13(y-7)^2-13(x+6)^2=52

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

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Example Question #1 : Hyperbolic Functions

Which of the following answers best represent $2\sinh(x)$ ?

Possible Answers:

e^x+e^-^x

e^x-e^-^x

$\frac{e^x+e^-^x}{2}$

$\frac{e^x-e^-^x}{2}$

$\frac{2}{e^x+e^-^x}$

Correct answer:

e^x-e^-^x

Explanation:

The correct definition of hyperbolic sine is:

$\sinh(x)= \frac{e^x-e^-^x}{2}$

Therefore, by multiplying 2 by both sides we get the following answer,

$2\sinh(x)=e^x-e^-^x$

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Example Question #5 : Hyperbolic Functions

Which of the following best represents $\cosh(x)-\sinh(x)$ , if the value of is zero?

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{2}$

Correct answer:

Explanation:

Find the values of hyperbolic sine and cosine when x is zero. According to the properties:

$\sinh(0)=\frac{e^x-e^{-x}}{2}=\frac{e^0-e^{-0}}{2}=\frac{1-1}{2}=0$

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

Therefore:

$\cosh(0)-\sinh(0)=1-0=1$

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Example Question #6 : Hyperbolic Functions

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

Possible Answers:

$\frac{1}{2}$

$\text{Does not exist}$

Correct answer:

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:

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

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Algebra II : Functions and Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #42 : Circle Functions

Example Question #491 : Functions And Graphs

Example Question #91 : Quadratic Functions

Example Question #492 : Functions And Graphs

Example Question #1 : Hyperbolic Functions

Example Question #1 : Hyperbolic Functions

Example Question #3 : Hyperbolic Functions

Example Question #1 : Hyperbolic Functions

Example Question #5 : Hyperbolic Functions

Example Question #6 : Hyperbolic Functions

All Algebra II Resources

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