how can I do this?

what is the equation of the graph?

Please help

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- Feb 20th 2007, 08:52 PMAmyGraph
how can I do this?

what is the equation of the graph?

Please help - Feb 20th 2007, 09:23 PMJhevon

Okay, so seeing two graphs on one set of axis suggests a piece wise function. The sharp turn in the lower graph indicates that it is an absolute value function. Now, I can't see the markings on the graph so clearly, so there may be some errors. I assume the lowest point on the graph, that is, where the sharp turn is is (-1,1), and that the y-intercept is 2 and that the horizontal graph is in line with y=5 and that the dots are above x=-3. So here goes:

An absolute valued function always gives positive answers, so whenever the answer is going to become negative, it it flips the graph upward, which is why we see a sharp turn. We see the graph flips when x is at -1, which means if x gets any smaller, the value will be negative, so then the absolute value part of the function is |x+1|, when x=-1, it becomes zero, anyless than -1, it is negative and therefore the graph flips up making a sharp turn. but we're not done yet. The y-intercept is 2. The y-intercept occurs when x is zero. In this case, if x=0, we have y=|1|=1. So to correct this, we just shift the graph upward by 1 by adding a constant 1, so the absolute value part of the function is |x+1| + 1.

The other part of the graph is easy (provided I can see it correctly), the graph is a constant 5, so it's y=5. Now we put these 2 functions together and obtain: - Feb 20th 2007, 09:24 PMJhevon
Whoops, there should be a pic there, but the forum tells me it's file size is too large. I'll try again

- Feb 20th 2007, 09:26 PMJhevon
okay, here it is. We see the circle attached to the end of the y=5 curve is unshaded, and therefore it means x is not equal to -3 at the point. The shaded circle at the end of the absolute value part says it can be equal.

- Feb 20th 2007, 09:41 PMAmythanks
thanks for your explanation.:) I understood this much very well. let me try to do the questions. can you do the last one (continuity) please

- Feb 20th 2007, 09:55 PMJhevon
No, the function is not continuous at x= -3.

Explanation:

A function f is continuous at a number a if

lim{x-->a}f(x)=f(a).

Here, we actually can find a value for f(a), that is f(-3)=2, but it is not equal to the limit. Since this lim at x=-3 does not exist. For a limit to exist, the left hand limit must be equal to the right hand limit. here, the left hand limit is 5, the right hand limit is 2 (looks like i ended up doing parts (a) and (b) for you by explaining this), so the limit does not exist and therefore, the function is not continuous.

Note: left hand limit is lim{x-->a-} and right hand limit is lim{x-->a+}

So you se, all the parts leading up to (e) was to give you clues. for the function to be continuous, we had to have the answers (a)=(b)=(c)=(d), but that was not the case. (a) was not equal to (b) and (c) didn't even exist