Graphing Rational Functions: Understanding Quadrant Placement and Behavior

[CanadianEh]

Question:
Consider the graph of f(x)=g(x)/h(x) where g(x) = 2^x and h(x)=x

a) Explain why it makes sense that the graph is in only the 1st and 3rd quadrants
b) Explain the behaviour of the curve in the first quadrant by making reference to the original functions.


I am sooo bad at these communication questions and I desperately need help!

 

[rock.freak667]

For the quadrants, consider the signs of the x values with the corresponding sign of the y values.

For example, in the first quadrant, is x-positive or negative? and is y positive or negative?

 

[CanadianEh]

In QI, the x and y values of g(x) and h(x) in the first quadrant are positive.
The x and y values of h(x) in QIII are all negative, but g(x) never goes into QIII because of the asymptote. How do I explain that? lol.

 

[rock.freak667]

The x and y values of h(x) in QIII are all negative, but g(x) never goes into QIII because of the asymptote. How do I explain that? lol.

For all values of x, 2^x is positive. So for 2^x/x and x is negative you will have positive/negative which gives a negative number, which would correspond to the sign of the y values in quadrant 3

 

[CanadianEh]

Thanks a lot, that makes a lot more sense. Can you help me with the end behaviours questions?

 

[rock.freak667]

Thanks a lot, that makes a lot more sense. Can you help me with the end behaviours questions?

In each quadrant, describe what happens as x becomes increasingly positive or negative as the case may be. Explain if there would be any significant features at any spefic value(s) of x and such.

 

[CanadianEh]

Well, in QI, as x gets bigger, 2^x quickly gets closer toward infiniti. Similarly, f(x)=x also gets closer to infinite, but at a slower rate. In QIII, f(x)=x goes to negative infiniti.

 

[rock.freak667]

Well, in QI, as x gets bigger, 2^x quickly gets closer toward infiniti. Similarly, f(x)=x also gets closer to infinite, but at a slower rate.

Right so f(x) will tend to infinity.

In QIII, f(x)=x goes to negative infiniti.

so as x goes to negative infinity 2^x will tend to what value?

 

[CanadianEh]

Zero, right?

 

[rock.freak667]

Zero, right?

yes. So would 2^x reach zero before x reaches negative infinity or does x reach before 2^x?

 

[CanadianEh]

2^x would reach zero before x reaches negative infiniti, correct?

 

[rock.freak667]

2^x would reach zero before x reaches negative infiniti, correct?

good good...anthing special happens when x goes to zero?

 

[CanadianEh]

are you referring to the asymptote?

 

[Chrisas]

b) Explain the behaviour of the curve in the first quadrant by making reference to the original functions.

Although it's good practice (and fun too) to consider both quadrants, just thought I'd point out that the problem only needs an answer for first quadrant for part b).

 

[CanadianEh]

Yeah, sorry I got side tracked. I'm looking at the two parent functions and I can't figure out how to explain the new end behaviours of the curve in QI. :S

 

[Chrisas]

It appears you and rock did a pretty good job of this already. You just need to put it all together and summarize. You found the asymptote as x goes to zero by considering what g(x) and h(x) does, and you described the fact the the function blows up at x towards infinity by considering how fast the numerator g(x) blows up compared to h(x). Just need to put it down on paper.

Now, some other things you might want to write. Does the Quadrant I part of the function stay within Quadrant I? Or does it cross the x-axis at some point? How do you know? (You already know it doesn't cross the y-axis because it blows up there.)

If it goes to positive infinity at x=0 and blows up at x=Inf, what does that imply that the function has at least one of in the middle? (Hint: think about minimums and maximums in the curve)

 

[CanadianEh]

It appears you and rock did a pretty good job of this already. You just need to put it all together and summarize. You found the asymptote as x goes to zero by considering what g(x) and h(x) does, and you described the fact the the function blows up at x towards infinity by considering how fast the numerator g(x) blows up compared to h(x). Just need to put it down on paper.

Now, some other things you might want to write. Does the Quadrant I part of the function stay within Quadrant I? Or does it cross the x-axis at some point? How do you know? (You already know it doesn't cross the y-axis because it blows up there.)

If it goes to positive infinity at x=0 and blows up at x=Inf, what does that imply that the function has at least one of in the middle? (Hint: think about minimums and maximums in the curve)

Yeah, that part of the curve is all in QI because you can clearly see that both the end behaviours are going to positive infiniti. As for the second part, does that imply that there's at least one asymptote?

 

[Chrisas]

Yeah, that part of the curve is all in QI because you can clearly see that both the end behaviours are going to positive infiniti.

That's not good enough to say that. It could fall from positive infinity at x=0, fall past the x-axis line to negative y, then turn around and head back up, cross the x-axis line back to positive and then go to positive infinity at x=infinity. As Rock pointed out, one way to know that doesn't happen is that g(x) and h(x) both stay positive so the result has to stay positive and never goes negative.

However, I'm saying there is a second feature of the graph that you could also talk about. What must happen to the numerator, g(x), if the function were to cross the x-axis? What happens to g(x) at the moment the function just crosses. However, Rock's answer showing that it is always positive is sufficient and you could ignore this part if you want.

As for the second part, does that imply that there's at least one asymptote?

No, that's not what I was fishing for. I was asking about function minimums and maximums. I don't know if you've gone that far in your course yet. It's probably not necessary to consider it.

 

[CanadianEh]

Well, isn't the reason why it shoots up to positive inifiniti near x=0 because it's 2^x divided by a really small X value from the f(x)=x curve? Like, when you divide a a number by a tiny number, it gives you a huge number, correct? And as for the other side of the curve, the 2^x function is still going up faster than f(x)=x which is why it continues in an exponential-esque shape. Correct?

 

[Chrisas]

Yes.

But your stuck on looking at the ends. That may be fine for your problem. What I was trying to get you to consider is what happens in the middle.

 

[CanadianEh]

Ahhh, I think I get it now. I think the reason why the middle is the way it is is because that is where the rate of change in both functions is similar.

 

[Mark44]

yes. So would 2^x reach zero before x reaches negative infinity or does x reach before 2^x?

I'm not sure I understand the question here. $\lim_{x \rightarrow -\infty}2^x = 0$ but 2^x is never zero for any finite value of x (no matter how negative). I might be misinterpreting your drift, though.

CanadianEh,
There is now, unfortunately, a word "infiniti" but it is an automobile brand. The word you are seeking is "infinity" -- with a y.

 

[CanadianEh]

I'm not sure I understand the question here. $\lim_{x \rightarrow -\infty}2^x = 0$ but 2^x is never zero for any finite value of x (no matter how negative). I might be misinterpreting your drift, though.

CanadianEh,
There is now, unfortunately, a word "infiniti" but it is an automobile brand. The word you are seeking is "infinity" -- with a y.

I think I figured it out. Thanks for everyone's help. Thanks for the correction Mark44! :)