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Thank you Bob, you hit the nail on the head. Now that school has reopened for teachers just yesterday I was able to go speak with him. While it took a bit of time as he didn’t pull any references and didn’t quite use clear terminology of “lines of symmetry” I got what he was saying and then googled “gradients of the lines of symmetry” and it became clear. As you said, once knowing this the answers then follow with calculation.
While I understand that he has taught this concept in grade 11, I feel that there is no clear indication on the diagram that the line f(x) IS a line of symmetry. While this exam may not have stated it, it is somewhat common knowledge and often stated in writing that “any diagrams” are not to scale and thus not to assume anything unless indicated by a symbol, statement or number.
As I tried to explain to him, while it “looked” like the line cut through the asymptotes at 45°, there was no proof of that. One could draw a line that perhaps was 40° - 44° that also cut through the two lines of the hyperbola which would not be a line of symmetry and thus the assumption that was suppose to be taken on the exam would then have been incorrect.
In terms of k if I recall correctly (and I will check my summary notes and get back to you) I have seen it before in formulas here in SA.
Terrific Bob, Thank you so much for the confirmation. I really appreciate it.
Thank you for your comments Bob and also about posting to imgur.com, I wasn't sure if you wanted me to use your link or simply put it up on imgur.com thus I did the later. Here is the link (hopefully it works):
https://imgur.com/a/AUbwRKR (it seemed to work by cutting and pasting it to my browser, as when I look at my preview this link doesn't show up like it does in your post)
Glad to hear about your thoughts on the maths teacher's comment regarding the gradients being -1 and +1, I too felt that simply looking at the hyperbola lines the gradients would be different from point to point on the lines.
Hopefully you can view the actual question at the link, in my opinion another point on the f(x) line needs to be provided in order to work out the answers to the questions...Let's see what you and others conclude.
The graphs of f(x) = mx + c and g(x) = [k/(x-b)] + c is shown below (not here as I don’t seem to be able to post a picture). A is the x-intercept of g. The asymptotes of g intersect at D (-1;-3). C is the y-intercept of both graphs. B is the x-intercept of f. The two graphs also intersect at E.
1) determine the equation of g.
2) determine the coordinates of A and B
3) determine the coordinates of E
4) value(s) of x for which: (a) g(x)<0 and (b) f(x) =< g(x)
To provide a bit of background - I am a Canadian who goes to South Africa each year for almost 7 months and I run free after school study groups at two high schools in disadvantaged areas in the town where I live.
I have gotten the handwritten answer sheet from the maths teacher, however it is not making sense. I am struggling to figure out m (gradient) for f(x). When I queried him on how he got m he said “the gradients of a hyperbolic function are -1 and 1, the points are the asymptotes. The line is passing through the point of intersection so you just take one, the positive 1 in this case. No calculations.”
I cannot find any reference to the above gradient statement and to me the gradient statement doesn’t seem to make sense, however I may be missing something.
Might someone shed some light on this problem with some explanation of reasoning in reaching the answers?
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