Question

How do you graph `y=4/(x-3)+2` using asymptotes, intercepts, end behavior?

Answer 1

Vertical asymptote at `x=3`, horizontal asymptote at
`y=2`, x intercept is at `(1,0)`, y intercept is at `(0,2/3)`.
End behavior : `x-> -oo , y-> 2 and x-> oo, y-> 2`

Explanation:

`y= 4/(x-3)+2`

Vertical asymptote at `x-3=0 or x =3`

When `x-> 3^-, y-> -oo and x-> 3^+, y-> oo`

Since denominator's degree is higher than that of numerator,

horizontal asymptote is at `y=0+2 or y=2`

y intercept is at `x=0 :. y= 4/(0-3)+2=-4/3+2 or y=2/3`

x intercept is at `y=0 :. 0 = 4/(x-3)+2 or -2 =4/(x-3)` or

`-2(x-3)=4 or 2 x = 6-4 or 2 x =2 or x=1`

End behavior : When `x-> -oo , y-> 2 and x-> oo, y-> 2`

graph{(4/(x-3))+2 [-40, 40, -20, 20]} [Ans]