After watching the 5 minute video linked below, answer the following questions:
1. The first function had an error message in the table at x = 0. Why?
2. Does this problem matter when "eye-balling" the limit?
3. If you were using your calculator to assist you in drawing a graph, what would you put on your graph that is not on the calculator screen?
4. The second function had error messages at x=1 and at x= -1. What is the difference between them?
5. If you're approaching a hole in the graph of a rational function, the limit still exists. How about if you're approaching a vertical asymptote?
Rational Functions - Use a calculator??
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1. the first function had an error at X=0 because X was the denominator. You have to break down the nominator so you can cancel out the X on the bottom.
ReplyDelete2. Yes, this matters when eyeballing the limit because you aren't able to just look at it and find the limit.
3. You would put a hole at (0,8)
4. There is a hole in the graph at X=1 whereas there is an asymptote at X=-1.
5. Yes, there still would be a limit. you would have to distinguish what way you are coming from because as X approaches it from one way, the graph goes to infinity and as you approach from the other way, the graph goes to negative infinity.
1. There is an error at x=0 for the first function because x is the denominator which would make it undefined at zero.
ReplyDelete2. It still matters because you still have to find the limit.
3. A hole at the point (0,8)
4. At x=-1 there is an asymptote and at x=1 there is a hole
5. There would still be a limit because you can approach the graph still from the left or the right
1. If you enter the function as shown, there would be an error message at x=0 because x is the denominator and whenever the denominator=0, the function is undefined. But if you simplify the function to y=x+8, you get the same graph and in your table, x=8.
ReplyDelete2. It matters because when you "eye-ball" the function you get errors on -1 and 1 but when you simplify the function to y=1/(x-1) the error only appears on 1.
3. As in the first function, I would place a hole at the point (0,8).
4. The asymptote is at x=-1 and the hole is at x=1. If the function was simplified, it would equal y=1/(x-1) the error only appears on 1.
5. The limit would still exist because you could approach it either way.
1. There is an error at x=0 because x is the denominator. The function is undefined wherever the denominator equals 0.
ReplyDelete2. It still matters because you still have to find the limit and accurately represent the whole when drawing the graph.
3. The whole is at (0,8).
4. The error message at x=-1 is an asymptote, whereas the error message at x=1 is a hole.
5. There would still be a limit because the point will still be approached either way.
1. a denominator of 0 would make an undefined function
ReplyDelete2. yes, you still have to identify the limit
3. I would put a hole at (0,8), separate from what the calculator displays.
4. x=-1 is an asymptote, but x=1 is a hole, not displayed by the calculator
5. the limit would still exist, regardless of which direction it is approached from
1. x is by itself in the denominator
ReplyDelete2. yes, because it doesnt tell you where the limit is on the graph
3. an open circle at (0,8)
4. when you factor (x^2-1) you get (x+1)(x-1), so there are two holes
5. The limit does not exist, because x approaches infinity from one side and negative infinity from the other side.
1. At x=0 for that first function, the value of 0 would render the answer undefined because of the numerator divided by a denominator of 0.
ReplyDelete2. It does matter because you can't spot a limit by executing the equation. Looking at the graph helps identify the limit.
3. A hole would be put at point (0,8) to show it wasn't possible because it was undefined.
4. When x approaches -1, it creates an asymptote. When x hits 1, it just creates a hole in the graph because it's undefined.
5. Yes, there is still a limit on vertical asymptotes because it would create values climbing to infinity and -infinity.
-Michael Ferrer, Period 6