In order to find the center and the vertices of the hyperbola given in the problem, we must examine the standard form of a hyperbola:

The point (h,k) gives the center of the hyperbola. We can see that the equation in this problem resembles the second option for standard form above, so right away we can see the center is at:

In the first option, where the x term is in front of the y term, the hyperbola opens left and right. In the second option, where the y term is in front of the x term, the hyperbola opens up and down. In either case, the distance tells how far above and below or to the left and right of the center the vertices of the hyperbola are. Our equation is in the first form, where the x term is first, so the hyperbola opens left and right, which means the vertices are a distance  to the left and right of the center. We can now calculate  by identifying it in our equation, and then go 3 units to the left and right of our center to find the following vertices:

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

, where  is the center of the hyperbola.

Start by putting the given equation into the standard form of the equation of a hyperbola.

Group the  terms together and  terms together.

Factor out  from the  terms and  from the  terms.

Complete the squares. Remember to add the amount amount to both sides of the equation!

Add  to both sides of the equation:

Divide both sides by .

Factor the two terms to get the standard form of the equation of a hyperbola.

The slopes of this hyperbola are given by the following:

For the hyperbola in question,  and .

Thus, the slopes for its asymptotes are .

Now, plug in the center of the hyperbola,  into the point-slope form of the equation of a line to get the equations of the asymptotes.

For the first equation, 

For the second equation,

is the standard form of a horizontal hyperbola with center . Set ; this hyperbola has its center at .

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.

Which of the following represents a horizontal transformation of v(t) 3 units to the right?

To perform a horizontal transformation on a function, we need to add or subtract a value within the function, which looks something like this:

Now, counter intuitively, when we shift right, we will subtract. If we wanted to shift left, we would add.

So, to shift 3 to the right, we need:

There are four fundamental transformations that allows us to think of a function  as a transformation of a function , 

In our case,  and , so the width and/or height of our function will not change in the coordinate plane. 

We have  and . The number  will shift the function up  units along the -axis on the coordinate plane. The number  will shift  unit to the right on the coordinate plane. 

The correct answer is . Inside the portion being squared the distance moved is opposite the sign and is horizontal. Outside the squared portion the distance moved follows the sign (plus is up and minus is down) and is vertical.

For example the incorrect answer  would have its vertex at (1,-5).

To shift a polynomial  to the right by 2, we must replace x with x-2 in whatever the expression for the polynomial is. The logic of this is that every x value has a y value associated with it, and we want to give every x value the y value associated with the point that is 2 before it.

So, to get our shifted polynomial, we plug in x-2 as noted.

and then we combine like terms:

As a general rule, if you have a function , then in order to reflect across the x-axis, we compute , and in order to reflect across the y-axis, we compute . In our case, we are asked to compute the latter.

So, if , then .

To compute a reflection about the x-axis, calculate , and to calculate a reflection about the y-axis, calculate . To compute a reflection about the origin, simply combine both reflections into .

In our case, .

So,