How do i find the asymptotes of:
y=(1/x-4)+3
did this for you three days ago.
http://www.jiskha.com/display.cgi?id=1456531698
still unclear?
Yes unclear
you cannot divide by zero.
so, when x=4, 1/(x-4) is undefined. There is a vertical asymptote there. For the horizontal asymptote, you need to see what y does when x gets huge. See the graph at
http://www.wolframalpha.com/input/?i=1%2F%28x-4%29%2B3
use google to find more examples and explanations.
To find the asymptotes of the function y = (1/(x-4)) + 3, you need to consider two types of asymptotes: vertical and horizontal.
1. Vertical Asymptotes:
Vertical asymptotes occur when the denominator of a fraction becomes zero. In this case, the denominator is x - 4. To find the vertical asymptote(s), set the denominator equal to zero and solve for x:
x - 4 = 0
x = 4
So, the vertical asymptote of the function is x = 4.
2. Horizontal Asymptotes:
Horizontal asymptotes occur when the function approaches a certain value as x approaches positive or negative infinity. To find the horizontal asymptote(s), observe the behavior of the function as x approaches infinity and negative infinity.
In this case, as x approaches positive infinity, the 1/(x-4) term approaches zero, since x becomes much larger than 4. Therefore, as x approaches infinity, the function approaches 3.
As x approaches negative infinity, the 1/(x-4) term approaches zero, since x becomes much smaller than 4. Therefore, as x approaches negative infinity, the function approaches 3.
Hence, there is a horizontal asymptote at y = 3.
To summarize:
- The vertical asymptote of the function y = (1/(x-4)) + 3 is x = 4.
- The horizontal asymptote of the function is y = 3.