Graphing Hyperbolic Functions - Algebra 2

To determine which direction a hyperbola opens, first get the equation into standard form for a conic section:

This equation gives us a hyperbola when the coefficient in front of either the x-squared or the y-squared term (but not both!) is negative. In this problem, the coefficient in front of the x-squared term is positive, but the coefficient of the y-squared term is negative. Here are the rules for hyperbola directions:

• If the coefficient of the x-squared term is positive but the coefficient of the y-squared term is negative, this is a hyperbola that opens horizontally.
• If the coefficient of the x-squared term is negative but the coefficient of the y-squared term is positive, this is a hyperbola that opens vertically.

Thus, we have a graph of a horizontal hyperbola.

The standard form of a hyperbola is

or the similar form with a positive term and negative term. So to start out getting this equation in standard form, we must complete the square on the quadratics in and .

the coefficient of is , so completing the square we get

and similarly with we get

and so our starting expression can be written as

Dividing by on both sides we get the standard representation of the hyperbola,

We start by noticing that our hyperbola is given in the following form:

In order to determine the vertices of the hyperbola, we must first locate its center. Using the standard form given above, we know the center of the hyperbola occurs at the point (h,k), so for the equation given in the problem the center is at (2,-3). Now that we know the location of the hyperbola's center, our next step is to determine how far the vertices are from the center of the hyperbola. Looking at our equation, we can see it is in the form where the x term occurs first, which means the hyperbola opens left and right as opposed to up and down (which would be the case if the y term occurred first. Given this information, we know the vertices of the hyperbola are going to be a distance to the left and right of the center. The denominator of the x term in the hyperbolic equation is 16, which means is equal to 4, so the vertices of the hyperbola will be 4 units to the left and right of the center (2,-3), which gives us:

and

To determine which direction a hyperbola opens, first get the equation into standard form for a conic section:

This equation gives us a hyperbola when the coefficient in front of either the x-squared or the y-squared term (but not both!) is negative. In this problem, the coefficient in front of the x-squared term is positive, but the coefficient of the y-squared term is negative. Here are the rules for hyperbola directions:

• If the coefficient of the x-squared term is positive but the coefficient of the y-squared term is negative, this is a hyperbola that opens horizontally.
• If the coefficient of the x-squared term is negative but the coefficient of the y-squared term is positive, this is a hyperbola that opens vertically.

Thus, we have a graph of a horizontal hyperbola.

The first thing we need to find for this hyperbola is the center. This is simply the point where and both equal , which is . Since the term is the positive one, the hyperbola opens horizontally, which means we need to look at the denominator of that term.

The denominator is which is , so our vertices are , or

and .

The standard form of a hyperbola is

or the similar form with a positive term and negative term. So to start out getting this equation in standard form, we must complete the square on the quadratics in and .

the coefficient of is , so completing the square we get

and similarly with we get

and so our starting expression can be written as

Dividing by on both sides we get the standard representation of the hyperbola,

We start by noticing that our hyperbola is given in the following form:

In order to determine the vertices of the hyperbola, we must first locate its center. Using the standard form given above, we know the center of the hyperbola occurs at the point (h,k), so for the equation given in the problem the center is at (2,-3). Now that we know the location of the hyperbola's center, our next step is to determine how far the vertices are from the center of the hyperbola. Looking at our equation, we can see it is in the form where the x term occurs first, which means the hyperbola opens left and right as opposed to up and down (which would be the case if the y term occurred first. Given this information, we know the vertices of the hyperbola are going to be a distance to the left and right of the center. The denominator of the x term in the hyperbolic equation is 16, which means is equal to 4, so the vertices of the hyperbola will be 4 units to the left and right of the center (2,-3), which gives us:

and

The first thing we need to find for this hyperbola is the center. This is simply the point where and both equal , which is . Since the term is the positive one, the hyperbola opens horizontally, which means we need to look at the denominator of that term.

The denominator is which is , so our vertices are , or

and .

To determine which direction a hyperbola opens, first get the equation into standard form for a conic section:

This equation gives us a hyperbola when the coefficient in front of either the x-squared or the y-squared term (but not both!) is negative. In this problem, the coefficient in front of the x-squared term is positive, but the coefficient of the y-squared term is negative. Here are the rules for hyperbola directions:

• If the coefficient of the x-squared term is positive but the coefficient of the y-squared term is negative, this is a hyperbola that opens horizontally.
• If the coefficient of the x-squared term is negative but the coefficient of the y-squared term is positive, this is a hyperbola that opens vertically.

Thus, we have a graph of a horizontal hyperbola.

The standard form of a hyperbola is

or the similar form with a positive term and negative term. So to start out getting this equation in standard form, we must complete the square on the quadratics in and .

the coefficient of is , so completing the square we get

and similarly with we get

and so our starting expression can be written as

Dividing by on both sides we get the standard representation of the hyperbola,

We start by noticing that our hyperbola is given in the following form:

In order to determine the vertices of the hyperbola, we must first locate its center. Using the standard form given above, we know the center of the hyperbola occurs at the point (h,k), so for the equation given in the problem the center is at (2,-3). Now that we know the location of the hyperbola's center, our next step is to determine how far the vertices are from the center of the hyperbola. Looking at our equation, we can see it is in the form where the x term occurs first, which means the hyperbola opens left and right as opposed to up and down (which would be the case if the y term occurred first. Given this information, we know the vertices of the hyperbola are going to be a distance to the left and right of the center. The denominator of the x term in the hyperbolic equation is 16, which means is equal to 4, so the vertices of the hyperbola will be 4 units to the left and right of the center (2,-3), which gives us:

and

To determine which direction a hyperbola opens, first get the equation into standard form for a conic section:

This equation gives us a hyperbola when the coefficient in front of either the x-squared or the y-squared term (but not both!) is negative. In this problem, the coefficient in front of the x-squared term is positive, but the coefficient of the y-squared term is negative. Here are the rules for hyperbola directions:

• If the coefficient of the x-squared term is positive but the coefficient of the y-squared term is negative, this is a hyperbola that opens horizontally.
• If the coefficient of the x-squared term is negative but the coefficient of the y-squared term is positive, this is a hyperbola that opens vertically.

Thus, we have a graph of a horizontal hyperbola.

The first thing we need to find for this hyperbola is the center. This is simply the point where and both equal , which is . Since the term is the positive one, the hyperbola opens horizontally, which means we need to look at the denominator of that term.

The denominator is which is , so our vertices are , or

and .

The standard form of a hyperbola is

or the similar form with a positive term and negative term. So to start out getting this equation in standard form, we must complete the square on the quadratics in and .

the coefficient of is , so completing the square we get

and similarly with we get

and so our starting expression can be written as

Dividing by on both sides we get the standard representation of the hyperbola,

We start by noticing that our hyperbola is given in the following form:

In order to determine the vertices of the hyperbola, we must first locate its center. Using the standard form given above, we know the center of the hyperbola occurs at the point (h,k), so for the equation given in the problem the center is at (2,-3). Now that we know the location of the hyperbola's center, our next step is to determine how far the vertices are from the center of the hyperbola. Looking at our equation, we can see it is in the form where the x term occurs first, which means the hyperbola opens left and right as opposed to up and down (which would be the case if the y term occurred first. Given this information, we know the vertices of the hyperbola are going to be a distance to the left and right of the center. The denominator of the x term in the hyperbolic equation is 16, which means is equal to 4, so the vertices of the hyperbola will be 4 units to the left and right of the center (2,-3), which gives us:

and