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

1 Answer
Binayaka C.
Jul 13, 2018

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]

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