To put this into standard form, we will have to complete the square for both x and y. To do this, it will be really helpful to re-group our terms:

Factor the -2 out of the y terms:

Adding 4 completes the square for x, since

Adding 9 completes the square for y, since

We're adding this inside the parentheses, so really we are adding to both sides

re-write the left side with squared binomials, and add the numbers on the right side

divide both sides by 17

To write this in standard form, first we will have to complete the square for both x and y. It will be helpful to re-group the terms:

 Factor out the coefficients for both x squared and y squared

Adding 1 completes the square for x since

Since we're adding it inside those parentheses, we're really adding

Adding 16 completes the square for y since

Since we're adding it inside those parentheses, we're really adding

divide both sides by 6

To write this in standard form, first complete the square for x and y. It will be helpful to re-group our terms:

factor out the coefficients on the squared terms

Adding 9 will complete the square for x since

We're adding it in the parentheses, so really we're adding to both sides

Adding 1 will complete the square for y since

We're really adding to both sides

divide both sides by 36

The given clue is that the hyperbola will open either up or down.  This indicates that the regular  and  variables are swapped.  

If the hyperbola opens left or right, the equation of the hyperbola is:

Write the form of a hyperbola in the second form.

Substitute the center .

The hyperbola must minimally be in this form.  

The correct answer is:  

Re-write the equation so that the x terms are next to each other:

factor our 4 from the x terms, and add 20 to both sides

complete the square for x by adding 1 inside the parentheses. To keep both sides equal, add 4 to the right side as well, since the 4 is really being distributed to the 1 we added on the left.

re-write the x terms as a binomial squared

divide both sides by 24

To be written in standard form, the equation of a hyperbola must look like one of the following:

     OR     

Our equation looks most like the one on the left, since the x term is first.  To obtain standard form, we must have 1 on the right side of the equation.  To do this, we divide both sides of the equation by 128. 

Now, we can simplify the fractions to obtain standard form

Let us first remember what each part of the equation for a hyperbola in standard form means:

The point (h,k) gives the center of the hyperbola. 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. For each of these standard form expressions, respectively, the equations for the asymptotes are:

So we can see that the equations are essentially the same, except that the slope is used when the hyperbola opens left and right, and its reciprocal, is used when the hyperbola opens up and down. We can see in the hyperbola for our problem that the x term appears first, so it opens left and right, which means we use the first option above for the equations of the asymptotes. Comparing the standard form equation with that given in the problem, we can determine h, k, a, and b:

Now we now all of the values required for the asymptote formula, so we plug them in and get:

So our asymptotes are described by the following two equations:

Remember that the standard form for the equation of a hyperbola is expressed in one of the following two ways:

Where a hyperbola of the first form would open left and right, and a hyperbola of the second form would open up and down. The formula for the asymptotes of either form are given below, respectively:

We can see that the equation given in the problem follows the second format of standard form, so we'll use the second equation above for finding the asymptotes of our hyperbola, which opens up and down. Looking at the equation for our hyperbola, we can see that h=-3, k= 5, a=4, and b=3. We have all the values we need, so we'll simply plug them into the formula for the asymptotes:

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.

The equations of the asymptotes for this hyperbola are given by the following equations:

For the hyperbola in question,  and .

Thus, the equations for its asymptotes are 

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 the asymptotes for this hyperbola are given by the following:

For the hyperbola in question,  and .

Thus, the slopes for its asymptotes are .

Now, use the center of the hyperbola, , to plug into the point-slope form of a line to find the equations of the asymptotes.

The first asymptote has the following equation:

The second asymptote has the following equation: