Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1.

The vertices are at (3, 0) and (-3, 0).

The foci are at (5, 0) and (-5, 0).

Step-by-step explanation:

Compare given equation x^2/9 - y^2/16 = 1

against standard hyperbola form (x - h) ^2 / a^2 - (y - k) ^2 / b^2 = 1,

the hyperbola opens left and right with its center (h, k) at (0, 0).

a^2 = 9 so a = + / -3

The vertices are at (3, 0) and (-3, 0).

a^2 + b^2 = c^2 where c is 2x distance between foci.

9 + 16 = c^2

c = + / -5

The foci are at (5, 0) and (-5, 0).

Step-by-step explanation:

equation x squared over nine minus y squared over sixteen = 1 is the same as

x^2/9 - y^2/16 = 1

x is at its min/max when y^2=0, y=0

x^2/9 = 1

x = + 3

vertices (+3, 0)

c^2 = 9 + 16 = 25

c = + 5

foci (+5, 0)