Question

I have a graph of a function with equation; y=ln(ax+b). It has a vertical asymptote at x=1 and x-intercept at (4,0). Can you remind me of an efficient way to find parameters 'a' and 'b' please? I was thinking of relating it to transformations but it seems too long a method. Thanks.

Answer 1

ok ln[a(x-h)] shifts graph "h" units left/right

where x = h is vertical asymptote

we are given that h=1
--> y = ln[a(x-1)]

now plug in x-intercept (4,0)
0 = ln(3a) ---> 3a = 1 ---> a = 1/3

y = ln (1/3x - 1/3)
a = 1/3
b = -1/3

Answer 2

great stuff - thanks mate! You explained it really nicely.

Answer 3

Vertical retricemtote means undefined. You know that ln(0) = undefined right? and ln(ax+b). In other words
a+b = 0

Now the other equatioin, you know that ln(1) = 0 so it would generate (4,0) so ln(1) = ln(ax+b), conclude that
ax+b = 1 . Now plug in x=4
4a +b = 1

Now you have 2 equations:
4a + b = 1
a+ b= 0

b= -a
substitute

4a -a =1
a = 1/3

a+ b = 0
1/3 + b = 0
b = -1/3

Answer 4

that was a good response too. Thank you too Study100. I know this stuff but I am out of practice so your help is much appreciated.