# How do you find the foci and sketch the hyperbola #x^2/9-y^2/4=1#?

We know that the standard Cartesian form for the equation of a hyperbola with a transverse horizontal axis,

has foci at

If we write the given equation in the same form as equation [1], then it is a simple matter to find the foci:

Now that we have the given equation in the same form, the computation for the foci is trivial:

To graph the equation you will need the vertices:

And you will need the equations of the asymptotes:

Here is a graph of, the equation, the foci, the vertices, and the asymptotes.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- How do you find the center, vertices, foci and asymptotes of # x^2/7 - y^2/9=1#?
- How do you write the standard form of the hyperbola #-x^2+y^2-18x-14y-132=0#?
- How do you write a standard form equation for the hyperbola with foci are (-6,0) and (6,0) and the difference of the focal radii is 10?
- How do you find the standard form of the equation of the hyperbola given the properties foci #(+-5,0)#, length of the conjugate axis 6?
- How do I graph the hyperbola with the equation #4x^2−25y^2−50y−125=0#?

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