Confused with maths problem?? Help (limits and functions)

DarkDusk

Hi,
Long story short - Watching videos on Khan Academy to help ease myself into LC HL maths. Going through the "Pre-Calculus" playlist which begins with limits. The problem I'm having trouble with is located here. I know how he gets the answer and blah blah blah BUT I haven't a clue how to graph the function. The function is: x^2 - 6x + 9 (as you can see in the video)
x^2 - 9
The answer at the end of the video which I understand is 0. I plotted the top line of the function and it shows that y=0 when x=3. So what's the point of the bottom line??? This is what confuses me. Could the problem not just be written without the bottom line (x^2 - 9)??? Help would be greatly appreciated.

Thank you,
James.

Turtwig

The "bottom line" makes your function completely different. Perhaps it might help to think of the function as :

[latex] f(x) = \frac{x^2}{x^2-9}-\frac{6x}{x^2-9}+\frac{9}{x^2-9}[/latex]


Graphs.png
The blue curve is [latex]f(x)=x^2-6x+9[/latex]
The red curve is [latex]f(x)=\frac{x^2-6x+9}{x^2-9}[/latex]

sponsoredwalk

Since the degree of the numerator is one higher than the degree of the
denominator this function will have a slanted (oblique) asymptote. This
is an easy way to understand how to graph these things.

https://www.youtube.com/watch?v=W8ASTRfEMVo

Turtwig

sponsoredwalk wrote: »

Since the degree of the numerator is one higher than the degree of the
denominator this function will have a slanted (oblique) asymptote. This
is an easy way to understand how to graph these things.

??
Do they not have the same degree?

DarkDusk

Malty_T wrote: »

The "bottom line" makes your function completely different. Perhaps it might help to think of the function as :

[latex] f(x) = \frac{x^2}{x^2-9}-\frac{6x}{x^2-9}+\frac{9}{x^2-9}[/latex]


Graphs.png
The blue curve is [latex]f(x)=x^2-6x+9[/latex]
The red curve is [latex]f(x)=\frac{x^2-6x+9}{x^2-9}[/latex]

Thank you, this helped a lot!! :):)

sponsoredwalk

Good god! Don't know how I missed that! Honestly, honestly honestly...
Was pure bias, seeing what I wanted to see

Find the vertical & horizontal asymptotes, find any holes & plug in a few
test points. Use the method in that video, it's the same idea :cool:

Michael Collins

Yes, the answer would be the same - but it wouldn't be the same problem. Obviously, as others have said, the function (and hence its graph) is different when the denominator is changes, but not only this - if the bottom line were left out the limit problem would be a good bit easier. Let me explain:

If the bottom were left out you'd just have

[latex] \displaystyle f(x) = x^2 - 6x + 9 [/latex]

the limit of which as x approaches 3 is easy to get (because this function is continuous) i.e. you just sub in x = 3, and get 0. Easy.

But with the denominator included you have a good bit of extra work to do, because now the function isn't defined at x = 3, but the limit can still exist as you approach 3 - in this case it happens to equal 0 also. This is not always the case; what's the answer to this problem, for example

[latex] \displaystyle \lim_{x \to 3} \frac{18(x-3)}{x^2-9}.[/latex]

Limits which end up as 0/0 when you try to sub in the value are called indeterminte forms . They tend to come up a lot. So this question was selected to teach you this concept. It has introduced you to the idea of limits of non-continuous functions.

DarkDusk

Michael Collins wrote: »

Yes, the answer would be the same - but it wouldn't be the same problem. Obviously, as others have said, the function (and hence its graph) is different when the denominator is changes, but not only this - if the bottom line were left out the limit problem would be a good bit easier. Let me explain:

If the bottom were left out you'd just have

[latex] \displaystyle f(x) = x^2 - 6x + 9 [/latex]

the limit of which as x approaches 3 is easy to get (because this function is continuous) i.e. you just sub in x = 3, and get 0. Easy.

But with the denominator included you have a good bit of extra work to do, because now the function isn't defined at x = 3, but the limit can still exist as you approach 3 - in this case it happens to equal 0 also. This is not always the case; what's the answer to this problem, for example

[latex] \displaystyle \lim_{x \to 3} \frac{18(x-3)}{x^2-9}.[/latex]

Limits which end up as 0/0 when you try to sub in the value are called indeterminte forms . They tend to come up a lot. So this question was selected to teach you this concept. It has introduced you to the idea of limits of non-continuous functions.

Thank you too! Very helpful!:)

DarkDusk

Oh and... Just wondering if much of this is on the leaving cert HL course for maths. I'm just after finishing the JC and thought I'd geta bit of headstart on the LC course (I love maths ). One more question: If this is on the course is it one of the easier or harder parts.

Thanks,
James.

Turtwig

DarkDusk wrote: »

Oh and... Just wondering if much of this is on the leaving cert HL course for maths. I'm just after finishing the JC and thought I'd geta bit of headstart on the LC course (I love maths ). One more question: If this is on the course is it one of the easier or harder parts.

Thanks,
James.

In all honesty, difficulty can be subjective. If I tell you a part is regarded as hard, then you're actually more likely to think it is so even if it is actually what you would have previously regarded as easy.* Try not to think about how others appreciate a topic, just focus on it yourself. At the end of the day all that will matter is your result. So to answer your question, everything is easy once you understand it. Somethings will just take longer than others to understand. Somethings, will just take shorter to understand. The notion that you either have a maths ability or not is rubbish. Any problems you may have ask your teacher, ask here, ask a friend; just don't skip it.


*Conversely, if I were to tell you it is easy and you would have previously thought it was hard, you would be more likely to underestimate your own ability and give up a lot quicker.