Respuesta :
Think of asy. as limiting fences to where your graph can travel. If, for example, you graph y = 1/x properly, you'll see that the graph never crosses either the x- or the y-axis. As x increases, your graph will get closer and closer to the line y=0 (which happens to be the horiz. axis), but will not cross it. Similarly, as x approaches x=0, the graph gets closer and closer to the vert. axis, x=0, but will not cross it. Do you see how the asymptotes limit where the graph can go?
Vertical asy. stem only from rational functions and correspond to x-values for which the denominator = 0. As you know, we can NOT divide by zero. Instead, we draw a vertical line thru any x-value at which the rational function is not defined.
Horiz. asy. have to do with the behavior of functions as x grows increasingly large, whether pos. or neg. Go back and re-read my earlier comments on horiz. asy. As x grows incr. large, in the positive direction, the graph of y=1/x approaches, but does not touch or cross, the horiz. asy.I will stop here and encourage you to ask questions if any of this discussion is not clear.
Vertical asymptotes can be defined by looking for gaps in the domain. In the equation f(x) = 1/x, you cannot put 0 into the equation. This is because we can't divide by 0. Giving us a vertical asymptote.
Horizontal asymptotes can be found by using the holes in the domain to find the range values that are not possible. In the same equation as above, there is no way to get a value of y = 0. Thus, you have a horizontal asymptote.
Oblique asymptotes are a bit harder to find. For this you would have to have a higher degree polynomial in the numerator than the denomination. In this case you would use synthetic division to find the slant. Example: 5x^3/x^2 + 5.
Horizontal asymptotes can be found by using the holes in the domain to find the range values that are not possible. In the same equation as above, there is no way to get a value of y = 0. Thus, you have a horizontal asymptote.
Oblique asymptotes are a bit harder to find. For this you would have to have a higher degree polynomial in the numerator than the denomination. In this case you would use synthetic division to find the slant. Example: 5x^3/x^2 + 5.