Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**JanetBiggar****Member**- Registered: 2019-12-04
- Posts: 4

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?

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,763

hi Janet,

Welcome to the forum.

Tough to help without the diagram.

I have an on-line account at www.imgur.com and I can upload images there and then point to them in my posts using BBCode. The BBCode here is

`[img]https://i.imgur.com/xSC6dDg.gif[/img]`

but what you see is my diagram.

Hopefully it looks a bit like yours but I've had to guess some of the values on the graph.

The function g(x) has one vertical and one horizontal asymptote and this will be true whatever the values of k, b and c. Here's why:

If x is just bigger than b then the denominator of the fraction will be small and hence y will be large. As x tends towards the value b from the right y will tend to infinity. If x is just smaller than b, then the same thing happens except the fraction is now negative so as x approaches the value b from the left y tends to minus infinity. So one asymptote is the line x = b

As x tends to infinity the fraction reduces to zero so without the +c term the x axis is the asymptote. The curve tends to zero from above when x is positive (assuming k is also positive) and tends to zero from below when x is negative. So that would make the x axis the other asymptote (without the +c)

But we are told that the asymptotes cross at D (-1, -3) so I chose k = 7 (just a guess) then b = +1 (that is forced by the D information) and c = 4 (another guess)

So then I had to make a straight line. All I know is it goes through the same y intercept as g(x), which is 4 on my diagram. So I chose an arbitrary gradient to finish the diagram.

So where do we go from here? Hopefully you've got a similar diagram, but not quite the same. If you post back the changes I'll have a proper go at the diagram and then I can help you finish the question.

the gradients of a hyperbolic function are -1 and 1

No idea what that means. Hyperbolas have continuously changing gradients. And it's not the gradients of the asymptotes.

f(x) = mx + c and g(x) = [k/(x-b)] + c

Question uses 'c' for both functions. Tried that and it yields k = 0 which makes the hyperbola disappear altogether so I don't think the 'c' value can be the same for both.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

Offline

**JanetBiggar****Member**- Registered: 2019-12-04
- Posts: 4

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.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,763

hi Janet,

Thanks, I've got it now. I was close with my guessed version. I've tried a couple of values for k ( =1 and = 0.5) and here's the comparison:

Either could be the graph for g(x).

You're right. Not enough information; too many unknowns.

From the asymptote information we can conclude

If I let the y coordinate of C be p then

Eliminating p gives

and there doesn't appear to be any other clues to determine m and k.

k=1 looks most like the given graph, and then p = -2. But, as it stands, not enough to even answer the first part of the question.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

Offline

**JanetBiggar****Member**- Registered: 2019-12-04
- Posts: 4

Terrific Bob, Thank you so much for the confirmation. I really appreciate it.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,763

hi Janet,

I've had a further thought about this.

All hyperbolas have two lines of symmetry. This one is a translation (and possibly an enlargement depending on k) of y = 1/x

That standard hyperbola has axes of symmetry y = x and y = -x with gradients of 1 and -1. Thus our hyp will have the same. So if we know that BCDE is such an axis then we can see it has gradient 1 and all the answers follow fairly easily (but not without some calculation!) This would explain the earlier help you received.

But the wording of the question does not say that this line is a line of symmetry (or the major axis which amounts to the same thing). So why did your helper think it is? The diagram doesn't show the line as a line of symmetry either.

I have wondered all along why the questioner used k for one parameter. Why that letter? It makes sense for f(x) to have m and c. That's normal for straight lines. There are no equivalent letters to use for a hyperbola. So why not a, b and c? Why use k rather than a. It doesn't matter in terms of doing the question, but it strikes me as odd.

On the maths is fun teaching site there is a graph plotter (https://www.mathsisfun.com/data/function-grapher.php) that has an additional variable 'a'. This allows you to explore the effect of changing one variable on the resulting graph, easily just by moving the slider.

Perhaps the questioner had a similar grapher but one that used k rather than a. So perhaps they set up the hyperbola and experimented with different k values to get one that would make a decent question. Then forgot to change k to a number when the book went to print. You get a good question if you make k = 1 so that's my suggestion.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

Offline

**JanetBiggar****Member**- Registered: 2019-12-04
- Posts: 4

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.

Offline

Pages: **1**