1. ## Graph

how can I do this?
what is the equation of the graph?

2. Originally Posted by Amy
how can I do this?
what is the equation of the graph?

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:

3. Whoops, there should be a pic there, but the forum tells me it's file size is too large. I'll try again

4. 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.

5. ## thanks

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

6. 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