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 direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: