Graphs - College Algebra

Card 0 of 380

Question

Find the x and y intercepts for

Answer

To find the x or y intercept in any equation we set y=0 or x=0 respectively.

for the x intercept:

$\boldsymbol{x=6\pm \sqrt{7}}$

The y intercept does not exist because to solve this equation for y requires complex roots (there will be a negative sign under the radical).

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Question

Which of the following is true of the relation graphed above?

Answer

The relation graphed is a function, as it passes the vertical line test - no vertical line can pass through it more than once, as is demonstrated below:

Also, it is seen to be symmetric about the origin. Consequently, for each in the domain, - the function is odd.

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Question

Define $f(x) = \frac{10}{x- 1}$ .

Is this function even, odd, or neither?

Answer

A function is odd if and only if, for all , ; it is even if and only if, for all , . Therefore, to answer this question, determine by substituting for , and compare it to both and .

$f(x) = \frac{10}{x- 1}$

$f(-x) = \frac{10}{-x- 1}$

, so is not even.

$- f(x) = - \frac{10}{x- 1} = \frac{10}{-x+1}$

, so is not odd.

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Question

is an even function; .

True or false: It follows that .

Answer

A function is even if and only if, for all in its domain, . It follows that if , then

.

No restriction is placed on any other value as a result of this information, so the answer is false.

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Question

Find the focal points of the conic below:

$27x^{^{2}}+16y^{^{2}}+324x-96y+684 = 0$

Answer

The first thing we want to do is put the conic (an ellipse because the x2 and the y2 terms have the same sign) into a better form i.e.

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

where (h,k) is the center of our ellipse.

We will continue by completing the square for both the x and y binomials.

First we seperate them into two trinomials:

$\left ( 27x^2+324x+0\right )+\left ( 16y^2-96y+0\right )=-684$

then we pull a 27 out of the first one and a 16 out of the second

$27\left ( x^2+12x+0\right )+16\left ( y^2-6y+0\right )=-684$

then we add the correct constant to each trinomial (being sure to add the same amount to the other side of our equation.

$27\left ( x^2+12x+36\right )+16\left ( y^2-6y+9\right )=-684+27(36)+16(9)$

then we factor our trinomials and divide by 16 and 27 to get

$\frac{\left ( x+6\right )^2}{16}+\frac{\left ( y-3\right )^2}{27}=1$

so the center of our ellipse is (-6,3) and we calculate the distance from the focal points to the center by the equation:

$c^2=\left | a^2-b^2\right |$

and we know that our ellipse is stretched in the y direction because b>a so our focal points will be c displaced from our center.

with $c=\sqrt{11}$

our focal points are $\left ( -6,3+\sqrt{11}\right ) , \left (-6,3-\sqrt{11} \right )$

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Question

What is the equation of the elipse centered at the origin and passing through the point (5, 0) with major radius 5 and minor radius 3?

Answer

The equation of an ellipse is

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

where a is the horizontal radius, b is the vertical radius, and (h, k) is the center of the ellipse. In this case we are told that the center is at the origin, or (0,0), so both h and k equal 0. That brings us to:

$\frac{(x-0)^2}{a^2} + \frac{(y-0)^2}{b^2} = 1$

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

We are told about the major and minor radiuses, but the problem does not specify which one is horizontal and which one vertical. However it does tell us that the ellipse passes through the point (5, 0), which is in a horizontal line with the center, (0, 0). Therefore the horizontal radius is 5.

The vertical radius must then be 3. We can now plug these in:

$\frac{x^2}{5^2} + \frac{y^2}{3^2} = 1$

$\frac{x^2}{25} + \frac{y^2}{9} = 1$

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Question

An ellipse is centered at (-3, 2) and passes through the points (-3, 6) and (4, 2). Determine the equation of this eclipse.

Answer

The usual form for an ellipse is

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

where (h, k) is the center of the ellipse, a is the horizontal radius, and b is the vertical radius.

Plug in the coordinate pair:

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

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

Now we have to find the horizontal radius and the vertical radius. Let's compare points; we are told the ellipse passes through the point (-3, 6), which is vertically aligned with the center. Therefore the vertical radius is 4.

Similarly, the ellipse passes through the point (4, 2), which is horizontally aligned with the center. This means the horizontal radius must be 7.

Substitute:

$\frac{(x+3)^2}{7^2} + \frac{(y-2)^2}{4^2} = 1$

$\frac{(x+3)^2}{49} + \frac{(y-2)^2}{16} = 1$

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Question

Find the center of this ellipse:

$33x^2+13y^2-396x+26\sqrt{7}y+850=0$

Answer

To find the center of this ellipse we need to put it into a better form. We do this by rearranging our terms and completing the square for both our y and x terms.

$33(x^2-12x)+13(y^2+2\sqrt7y)=-850$

completing the square for both gives us this.

$33(x^2-12x+36)+13(y^2+2\sqrt7y+7)=429$

$33(x-6)^2+13(y+\sqrt7)^2=429$

we could divide by 429 but we have the information we need. The center of our ellipse is $(6,-\sqrt7 )$

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Question

Find the eccentricity of an ellipse with the following equation:

Answer

Start by putting this equation in the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where is the center of the ellipse.

Group the terms together and the terms together.

Factor out from the terms and from the terms.

Now, complete the squares. Remember to add the same amount to both sides!

Subtract from both sides.

Divide both sides by .

$\frac{(x^2-16x+64)}{4}+\frac{y^2-2y+1}{25}=1$

Factor both terms to get the standard form of the equation of an ellipse.

$\frac{(x-8)^2}{4}+\frac{(y+1)^2}{25}=1$

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, .

$e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}$

Next, find the distance from the center to the focus of the ellipse, . Recall that when , the major axis will lie along the -axis and be horizontal and that when , the major axis will lie along the -axis and be vertical.

is calculated using the following formula:

$c=\sqrt{a^2-b^2}$ for , or

$c=\sqrt{b^2-a^2}$ for

For the ellipse in question,

$c=\sqrt{25-4}=\sqrt{21}$

Now that we have found the distance from the center to the foci, we need to find the distance from the center to the vertex.

Because , the major axis for this ellipse is vertical. will be the distance from the center to the vertices.

For this ellipse, .

Now, plug in the distance from the center to the focus and the distance from the center to the vertex to find the eccentricity of this ellipse.

$e=\frac{\sqrt{21}}{5}$

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Question

Write the equation for an ellipse with center , foci $(1 \pm \sqrt{14}, 4)$ and a major axis with length 14.

Answer

The general equation for an ellipse is $\frac{(x-h)^2 }{a^2 } + \frac{(y-k)^2 }{b^2 } = 1$ , although if we consider a to be half the length of the major axis, a and b might switch depending on if the longer major axis is horizontal or vertical. This general equation has as the center, a as the length of half the major axis, and b as the length of half the minor axis.

Because the center of this ellipse is at and the foci are at $(1 \pm \sqrt{14} , 4 )$ , we can see that the foci are $\sqrt{14}$ away from the center, and they are on the horizontal axis. This means that the horizontal axis is the major axis, the one with length 14. Having a length of 14 means that half is 7, so . Since the foci are $\sqrt{14}$ away from the center, we know that $c = \sqrt{14}$ . We can solve for b using the equation :

$7^2 - \sqrt{14}^2 = b^2$

that's really as far as we need to solve.

Putting all this information into the equation gives:

$\frac{(x-1)^2 }{ 49} + \frac{(y - 4)^2 }{35 } = 1$

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Question

What is the equation of the ellipse given the following:

Vertices: (10,0), (-10,0)

Co-vertices: (0,7), (0,-7)

Answer

Standard equation of an ellipse: $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$

From the given information, the ellipse is centered around the origin (0,0), so h and k are both 0.

The coordinates of the vertices are on the x-axis, which is the major axis. The vertices are a units away from the center. Here, a = 10.

The coordinates of the co-vertices are on the y-axis, which is the minor axis. The co-vertices are b units away from the center. Here, b = 7.

$\frac{x^{2}}{100}+\frac{y^{2}}{49}=1$

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Question

The graph of the equation

$16x^{2} + 12y^{2} -96x-48y =0$

is an example of which conic section?

Answer

The quadratic coefficients in this general form of a conic equation are 16 and 12. They are of the same sign, making its graph, if it exists, an ellipse.

To determine whether this ellipse is horizontal or vertical, rewrite this equation in standard form

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

as follows:

Separate the and terms:

$16x^{2} + 12y^{2} -96x-48y =0$

$(16x^{2} -96x) + (12y^{2}-48y) =0$

Distribute out the quadratic coefficients:

$16 ( x^{2} -6x ) +12 ( y^{2}-4y) =0$

Complete the square within each quadratic expression by dividing each linear coefficient by 2 and squaring the quotient.

Since $\left (\frac{-6}{2} \right )^{2}= 9$ and $\left (\frac{-4}{2} \right )^{2}= 4$ , we get

$16 ( x^{2} -6x \textbf{+ 9} ) +12 ( y^{2}-4y\textbf{ + 4}) =0+ \textbf{?}+ \textbf{?}$

Balance this equation, adjusting for the distributed coefficients:

$16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =16(9)+ 12(4)$

$16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =192$

The perfect square trinomials are squares of binomials, by design; rewrite them as such:

$16 ( x-3 )^{2} +12 ( y -2 ) ^{2} =192$

Divide by 192:

$\frac{16 ( x-3 )^{2}}{192} +\frac{12 ( y -2 ) ^{2} }{192}=\frac{192}{192}$

$\frac{( x-3 )^{2}}{12} +\frac{( y -2 ) ^{2} }{16}=1$

The ellipse is now in standard form. $b^{2} > a^{2}$ , so the graph of the equation is a vertical ellipse

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Question

is an even function. Let $g(x) = \frac{f(x)}{x}$ .

Is an even function, an odd function, or neither?

Answer

A function is even if, for each in its domain,

.

It is odd if, for each in its domain,

.

Substitute for in the definition of :

$g(-x) = \frac{f(-x)}{-x} = - \frac{f(-x)}{x}$

Since is even, , so

$g(-x) = - \frac{f(x)}{x} = -g(x )$

This makes an odd function.

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Question

The graph of the equation

$16x^{2}+12y^{2}-96x+48y+384=0$

is an example of which conic section?

Answer

The quadratic coefficients in this general form of a conic equation are 16 and 12. They are of the same sign, making its graph, if it exists, an ellipse.

To determine whether this ellipse is horizontal or vertical, rewrite this equation in standard form

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

as follows:

Subtract 384 from both sides:

$16x^{2}+12y^{2}-96x+48y+384=0$

$16x^{2}+12y^{2}-96x+48y= -384$

Separate the and terms and group them:

$16x^{2} + 12y^{2} -96x-48y =-384$

$(16x^{2} -96x) + (12y^{2}-48y) =-384$

Distribute out the quadratic coefficients:

$16 ( x^{2} -6x ) +12 ( y^{2}-4y) =-384$

Complete the square within each quadratic expression by dividing each linear coefficient by 2 and squaring the quotient.

Since $\left (\frac{-6}{2} \right )^{2}= 9$ and $\left (\frac{-4}{2} \right )^{2}= 4$ , we get

$16 ( x^{2} -6x \textbf{+ 9} ) +12 ( y^{2}-4y\textbf{ + 4}) =-384+ \textbf{?}+ \textbf{?}$

Balance this equation, adjusting for the distributed coefficients:

$16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =-384+ 16(9)+ 12(4)$

$16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =-192$

The perfect square trinomials are squares of binomials, by design; rewrite them as such:

$16 ( x-3 )^{2} +12 ( y -2 ) ^{2} =-192$

Divide by :

$\frac{16 ( x-3 )^{2}}{-192} +\frac{12 ( y -2 ) ^{2} }{-192}=\frac{-192}{-192}$

$\frac{( x-3 )^{2}}{-12} +\frac{( y -2 ) ^{2} }{-16}=1$

Recall that the standard form of an ellipse is

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

This requires both denominators to be positive. In the standard form of the given equation, they are not. Therefore, the equation has no real ordered pairs as solutions, and it does not have a graph on the coordinate plane.

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Question

Give the foci of the ellipse of the equation

$\frac{x^{2}}{46}+ \frac{y^{2}}{19} = 1$

Round your coordinates to the nearest tenth, if applicable.

Answer

The equation of the ellipse is given in the standard form

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

This ellipse has its center at the origin . Also, since $a^{2} >b^{2}$ , it follows that the ellipse is horizontal. The foci are therefore along the horizontal axis of the ellipse; their coordinates are , where

$c = \sqrt{a^{2}-b^{2}}$

Substituting 46 and 19 for $a^{2}$ and $b^{2}$ , respectively,

$c = \sqrt{46-19} = \sqrt{27} \approx 5.2$ .

The foci are at the points and .

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Question

Give the eccentricity of the ellipse of the equation

$\frac{x^{2}}{25}+ \frac{y^{2}}{169} = 1$

Answer

This ellipse is in standard form

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

where $b^{2} > a^{2}$ . This is a vertical ellipse, whose foci are

$c= \sqrt{b^{2}-a^{2}}$

units from its center in a vertical direction.

The eccentricity of this ellipse can be calculated by taking the ratio $\frac{c}{b}$ , or, equivalently, $\frac{\sqrt{b^{2}-a^{2}}}{b}$ . Set $b^{2} = 169$ - making - and $a^{2} = 25$ . The eccentricity is calculated to be

$\frac{\sqrt{169-25}}{13} =\frac{ \sqrt{144}}{13} = \frac{12}{13}$ .

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Question

Give the foci of the ellipse of the equation

$\frac{(x- 2)^{2}}{64} + \frac{(y-6)^{2}}{49} = 1$ .

Round your coordinates to the nearest tenth, if applicable.

Answer

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

is the standard form of an ellipse with center . Also, since in the given equation, $a^{2} = 64$ and $b^{2} = 49$ - that is, $a^{2}> b^{2}$ , the ellipse is horizontal.

The foci of a horizontal ellipse are located at

$(h \pm c ,k)$ ,

where

$c= \sqrt{a^{2}-b^{2}} = \sqrt{64-49} = \sqrt{15} \approx 3.9$

Setting $h = 2, k = 6, c \approx 3.9$ , the foci are at

$(2 \pm 3.9, 6)$ , or

and .

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Question

Which of the following represents a vertical shift up 5 units of f(x)?

Answer

Which of the following represents a vertical shift up 5 units of f(x)?

A vertical translation can be accomplished by adding the desired amount onto the end of the equation. This means that f(x)+5 will shift f(x) up 5 units.

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Question

Using the information below, determine the equation of the hyperbola.

Foci: and

Eccentricity: $\frac{3}{2}$

Answer

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

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

Distance between foci =

Distance between vertices =

Eccentricity =

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 12, we can set that equal to .

$\small c= 6$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small c/a=3/2$

$\small \frac{c}{a}=\frac{3}{2}=\frac{6}{a}$

$\small 12=3a$

$\small a=4$

$\small \small a^{2}=16$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small 4^{2}+b^{2}=6^{2}$

$\small \small b^{2}=20$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 4, the center point will be on the same line. Hence, $\small k = 4$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 12, we can conclude that $\small h=1$

Center point: $\small \small (h, k)= (1,4)$

Thus, the equation of the hyperbola is:

$\small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{20}=1$

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Question

Using the information below, determine the equation of the hyperbola.

Foci: and

Eccentricity:

Answer

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

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

Distance between foci =

Distance between vertices =

Eccentricity =

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 8, we can set that equal to .

$\small 8=2c$

$\small \small c= 4$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small \small \small c/a=2$

$\small \small \frac{c}{a}=\frac{2}{1}=\frac{4}{a}$

$\small \small 4=2a$

$\small \small a=2$

$\small \small \small a^{2}=4$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small \small 2^{2}+b^{2}=4^{2}$

$\small \small \small b^{2}=12$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 8, the center point will be on the same line. Hence, $\small \small k = 8$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 8, we can conclude that $\small \small h=5$

Center point: $\small \small \small (h, k)= (5, 8 )$

Thus, the equation of the hyperbola is:

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

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