asymptotes and intercepts for f(x)= (2x+3)/(3x+4)?

Question

graph and find the inverse function and its graph
for precalcaulus

Answer 1

Asymptote at -4/3
You find this by looking at where the function is not defined. A function is not defined when the denominator is 0.
So when is 3x+4 = 0
==> 3x = -4
==> x = -4 /3

Y intercept at 3/4
To find y intercept, set x = 0, and solve for y
y = (2x+3) / (3x+4)
y = (2(0) + 3) / (3(0) + 4)
y = (0 + 3) / (0 + 4)
y = 3/4

X intercept at -3/2
To find x intercept , set y = 0, solve for x
0 = (2x+3) / (3x+4)
==>0*(3x+4) = 2x+3
==>0 = 2x+3
==>-3 = 2x
==>-3/2 = x

Inverse is y = (3-4x) / (-2+3x)
To find inverse, change all your y's in your function to x's and all x's to y's, then solve for y.
y = (2x+3) / (3x + 4)
Changes to
==> x = (2y+3) / (3y + 4)
==> 3xy + 4x = 2y + 3
==> -2y + 3xy + 4x = 3
==> y(-2+3x) + 4x = 3
==> y(-2+3x) = 3 - 4x
==> y = (3-4x) / (-2+3x)

Answer 2

Jeff's answer is correct, but he forgot to find the horizontal asymptote. The limit of f as x-> inf is 2/3. and this is also the limit as x -> -inf, so the horizontal asymptote is f = 2/3.