The difference between graphing the two equations is the first equation is a removable discontinuity and the second graph is a non removable discontinuity. To graph the first equation I first factored the top of the equation. After that I realized I was able to simplify the equation by canceling out the two x+1s. This creates a hole at -1 because -1+1 would be zero. After that I found the vertical asymptote a which would be -1 and 4, because 4-4=0. Since the degree of the numerator is greater than that of the denominator, there would be no horizontal asymptote. To make the numerator 0, you would have to plug in a 4 for x, that is where I got my x intercept. When you plug in 0 for x you get -4, that is where I got my y intercept. I then used all of these features to grt my graph laid out. I then created a table of x values within the asymptotes, and solved them to get the y values. Then I graphed the points.
For the second graph I started out the same way, except this time it was the denominator that needed to be factored. This time nothing needed to cancel out after I factored. The vertical asymptotes would be 2 and -2 since 2-2=0 and -2+2=0. The degrees are the same in this equation and 1/1 is 1, that's where the horizontal asymptote comes from. The intercept is 4 because 4-4=0 in the numerator. The y intercept ends up being 0 when you plug in 0 for x. Then I laid out the graph just as I did with the last equation and plotted the points the same way.
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