1 divided by 0 is undefined, how do i insert it on a graph?

Question

my points that i need to graph are x= -1, 0, 1, 2, 3, 4 and y= -0.3, -0.5, -1, undefined, 1, 0.5
How do I draw a curve graph using undefined? or did i do something wrong? the assignment was: For the function y= 1/x-2 give the y values for x=-1, 0, 1, 2, 3, 4. Using these values draw a graph.

Answer 1

The short answer is the graph is a hyperbole, which never actually crosses the y axis, therefore at x=0, y is undefined.

okay, first plot your points.

X | Y
------|------
-1 | -3
0 | Undefined
1 | -1
2 | -1.5
3 | -1.667
4 | -1.75

What you will find is two sets of curves, both approaching the Y axis, but never touching it. They are symmetric, so by plotting the positive X-axis, you can mirror the negative side fo the X-axis.

This is for the equation y=1/x-2.. y=1/(x-2) is a totally different graph, and is answered above.

Answer 2

Since when x = 2, y is undefined, there is a discontinuity at x =2 and so you can draw the line x=2 and it will be an asymptote. From the points given, it looks like you have a hyperbola which has the x-axis and the line x=2 as asymptotes. So the graph occurs in the 1st and 3rd and part of the 4th quadrant only. The graph is negative for - infinity

Answer 3

y = 1/(x - 2) has a vertical asymptote at x = 2, so you don't really "insert" it on a graph per say. You create a dashed vertical line at x = 2, and from x = 1, you curve downward to negative infinity, getting closer and closer to the asymptote x = 2 but not quite reaching it.

The same holds true from the right hand side, from x = 3 leftward; this curves upward, closer and closer to the asymptote straight up to infinity.

Answer 4

The answers above are right, but they didn't tell you what you wanted to know. To draw it on a graph, you put an open circle at the point that is undefined to show that that particular point is not included in the graph for this function, but all other points are.

Answer 5

You have a curve called a hyperbola. It resembles two different figures, which have limits called asymptotes. In your case, there is an asymptote at x=2. There are two parts to the curve. One part, the y-negative part, will curve down sharply, but never reach x=2 from the left. The other, the y-positive part, will "dive" down from x=2, to the positive y-points.

Answer 6

At 2, there will be a vertical asymptote. The graph exists only on each side of 2, not at 2.

Answer 7

Put in two points, like this:

(1.999999, -1e6)
(2.000001, +1e6)

I assume you meant y = 1 / (x-2)