So I have to find the x and y intercepts of f(x)=ln(x+1) from my understanding this means that the graph shifts on place to the left. In order to solve I set x+1 to equal 0 which is obviously x=-1. I get a little confused as to where I go next. Am I even on the right track?
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intercepts of ln function
- Thread starter sajoshi21
- Start date
So I have to find the x and y intercepts of f(x)=ln(x+1)
from my understanding this means that the graph shifts on place to the left. \(\displaystyle \ \ \ \) The graph of f(x) = ln(x + 1) is definitely shifted
one unit horizontally from the graph of f(x) = ln(x).
In order to solve I set x+1 to equal 0 which is obviously x=-1. \(\displaystyle \ \ \ \)That is incorrect for finding either an x-intercept or a y-intercept.
I get a little confused as to where I go next. Am I even on the right track?
For the x-intercept, let f(x) = 0 and solve for x. \(\displaystyle \ \ \ (?, \ 0)\)
For the y-intercept, evaluate the function at x = 0 (substitute x = 0). \(\displaystyle \ \ \ (0, \ ?)\)
You are very close to being on the right track.So I have to find the x and y intercepts of f(x)=ln(x+1) from my understanding this means that the graph shifts on place to the left. In order to solve I set x+1 to equal 0 which is obviously x=-1. I get a little confused as to where I go next. Am I even on the right track?
First, g(x) = f(x + a). If a is greater than zero, then the graph of g(x) is identical to the graph of f(x) except shifted a units to the left. If a is negative, then the shift is to the right. Let's see why. Suppose a is plus 1 so g(x) = f(x + 1). Further suppose that f(10) = 4.
So g(9) = f(9 + 1) = f(10) = 4. So do you understand this horizontal shift? You seem to, but perhaps this has dispelled any uncertainty you may have had. If you still do not understand, lett us know and we can give you some examples.
As lookagain said, finding x and y intercepts is a basic skill you learned long ago. At an x-intercept of f(x), f(x) = 0: f(x) touches the x-axis. If x is defined at 0, the y-intercept = f(0): f(x) touches the y-axis only where x = 0.
What is tricky about this problem is that ln(x) has no y-intercept because ln(x) is defined for only positive numbers. So ln(x + 1) is not defined at x = - 1. However, ln(x + 1) is defined at x = 0 because 0 + 1 = 1, which is a positive number.
Can you do the problem now?
If not either show us what you were able to do or explain where you are stuck.
So I do f(x)=ln(0+1)? that equals (1,0)? then f(0)=ln(x+1) this is where I get confused. Do I need to be paying attention to the ln at all or am I just solving for the intercepts. I'm so frustrated, sometimes math comes so easily and other times its like looking at a foreign language.
It IS a foreign language that amazingly is read and written by all numerate persons around the globe no matter what their natural language.So I do f(x)=ln(0+1)? that equals (1,0)? then f(0)=ln(x+1) this is where I get confused. Do I need to be paying attention to the ln at all or am I just solving for the intercepts. I'm so frustrated, sometimes math comes so easily and other times its like looking at a foreign language.
Solving for the intercepts of a function always involves the specific function. The x-intercept of h(x), if there are any, are those values of x where h(x) = 0. The y-intercept of h(x) is h(0).
f(x) = ln(x + 1). To find the y-intercept, f(0) = ln(0 + 1) = ln(1) = 0. That was fine, but the value of x involved is ZERO so the y-intercept
is (0, 0).
So where is f(x) = 0. Why we just found one x-intercept because f(0) = 0. The y-intercept is also an x-intercept. There is never more than one y-intercept, but there may be more than one x-intercept.
In the case of f(x) = ln(x + 1) is there more than one x-intercept? How can you be sure?