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Can a rational graph with a horizontal asymptote of y=0 still go through y=0?
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Sure it can, take f(x) = (2x+4)/(x^2 - 9) for example It has a zero at x = -2 and it still has a horizontal asymptote of y = 0
Thanks! But why can the line not cross the vertical asymptote but can the horizontal?
Because if the graph crossed the vertical asymptote, then it would mean that the x value is defined for that function But look at g(x) = 1/(x-3) It is undefined at x = 3, so there is a vertical asymptote there. If it crossed the vertical asymptote, then it would be defined at x = 3 (which is not true)
You can cross the horizontal asymptote as many times as you want, but you can never cross the vertical asymptote
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