Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y plus five squared divided by nine = 1.
a. vertices: (-5, 3), (-5, -5); foci: (-5, -6), (-5, 4)
b. vertices: (2, -5), (-4, -5); foci: (-4, -5), (2, -5)
c. vertices: (3, -5), (-5, -5); foci: (-6, -5), (4, -5)
d. vertices: (-5, 2), (-5, -4); foci: (-5, -4), (-5, 2)

C

Step-by-step explanation:

This hyperbola is a horizontal hyperbola of the standard form:

Since our equation is

,

a = 4 and b = 3.

The coordinates for the vertices are (±a, 0) and

the coordinates for the foci are (±c, 0).

We have a, but we need c.  To find c, we use Pythagorean's Theorem:

or

giving us that

c = 5.

But these a and c values have to be figured from the center of the hyperbola which is located at (-1, -5).  

For the vertices, then, we add the a value of 4 and -4 to the x value of the center, which is -1.  The -5 remains, since the vertices and the foci are on the same transcersal axis which is the line y = -5.

For the foci, then, we add the c value of 5 to -1, and again the -5 remains in the y position.

Vertices: (-1+4, -5)-->(3, -5) and (-1-4, -5)-->(-5, -5)

Foci: (-1+5, -5)-->(4, -5) and (-1-5, -5)-->(-6, -5)

Choice C

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^