How do you graph #y=-4/(x-6)-5# using asymptotes, intercepts, end behavior?

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Answer 1 · 2017-01-04T03:50:31

Vertical: #uarr x=6darr#.Horizontal: #larr y = -5 darr#.
x-intercept (y=0): 26/5. y-intercept (x=0); #-13/3#.

graph{(y+5)(x-6)((y+5)(x-6)+5)=0 [-25, 25, -12.5, 12.5]}

Ax+By+C= D/(Ex+Fy+G) represents the hyperbola

(Ax+By+C)(Ex+Fy+G)=D having the guiding asymptotes

Ax+By+C)(Ex+Fy+0 that meet at the center of the hyperbola.

Here, the form is

(y+5)(x-6)=-4.

The asymptotes are given by

(y+5)(x-6)=0

Separated, they are

the horizontal #y = -5# and the vertical x = 6.

The center is at the intersection (6, -5)

See the Socratic graph. The vertical asymptote is elusive in the plot,

.but understandable