SOLUTION: Show how to find the domain, range, x and y intercepts, and any asymptotes for G(x)= 1 + 2/(x-3)^2

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Question 1160: Show how to find the domain, range, x and y intercepts, and any asymptotes
for
G(x)= 1 + 2/(x-3)^2

Answer by khwang(438) About Me  (Show Source):
You can put this solution on YOUR website!
Since any denominator cannot be 0,
so x-3 cannot be 0 in y=G(x)= 1 + 2/(x-3)^2 and 3 should be
excluded from the domain of G.
Hence,the domain of G = R -{3} = (-oo,3) U(3,+oo)
Next,we see that 2/(x-3)^2 > 0 for all real x, so G(x) > 1+0 = 1.
So, the range of G is (1, +oo)

When x =0, G(0) = 1 + 2/9 = 11/9 is y-intercept.
Since y=G(x) >0 for all x, there are no x-intercepts of y=G(x).
The denominator in G(x) is x - 3= 0, which is a vertical asymptote
of G.
As x--> +oo, G(x) --> 1.So, a horizontal asymptote is y=1.