Mth 158
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I have several questions pertaining to chapter 5.
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Chapter 5 Behavior near a zero.
I am having trouble understanding the behavior near zero.
For example:
f(x) = x^2(x-4)(x+1)
I know the x-intercepts are -1, 0 and 4
But how dow the graph of f resemble that of the power function y = x^4
I am not seeing how the degree of 4 comes into this.
If you multiply out the factors x^2 ( x - 4) ( x + 1) you get x^2 ( x^2 - 3x - 4 ) = x^4 - 3 x^3 - 4 x^2.
When x is large, x^4 is much larger than -3 x^3 - 4 x^2.
For example, let x = 10. x^4 would be 10 000. -3 x^3 would be -3 000. -4 x^2 would be -400. So x^4 is already bigger than any of the other terms.
If x = 100 then x^4 = 100 000 000. -3x^3 would be -3 000 000, which is big but its magnitude is only 3% that of x^4. -4 x^2 would be -40 000, which is tiny compared to the value of x^4.
The bigger x gets, the less significant the terms -3 x^3 and -4 x^2 become.
So as x gets large, the graph of the polynomial differs less and less significantly from that of x^4.
The next question is finding the vertical, horizontal or oblique asymptotes of a rational function.
I can see how you find the vertical asymptotes, and I can follow along for a while on the others, but here is the part that confuses me:
R(x) = x - 12 / 4x^2 + x + 1
I know that the degree of the numerator is less than that of the denominator and that makes it proper. So we conclude that Y = 0 but the explanation of why y = 0 is not clear
Just as in the preceding explanation, as x gets large the value of the denominator differs less and less significantly from the value of just 4 x^2. The numerator differs from x only by -12, which becomes insignificant when x gets large.
As x gets large, then, the rational expression approaches x / (4 x^2), which is just 1 / (4 x). For large values of x, the value of 1/x is close to zero.
To convince yourself of this, substitute x = 1000 into the expression and evaluate it. Your answer will be very close to the value of just plain x / (4 x^2) , which is 1/4000 = .00025.
R(x) = x - 12 / 4x^2 + x + 1 is approx = to x / 4x^2 = 1/4x then it has an arrow pointing right and a 0 (the caption underneath says as x approaches - infinity or x approaches infinity) but what does that mean?
as x approaches - or infinity ....what? does it become 0, does Y become 0
x -> infinity means that x gets larger and larger, without any bound on how large. This means you can make x as big as you like, but you always have to think of what would happen as x gets even larger.
As x gets larger and larger, 1 / (4x) gets smaller and smaller. There is no limit to how large x can get, and there is no limit to how small 1 / (4x) can get.
So the limit as x approaches infinity of 1 / (4x) is zero.
This statement would be written
limit {x -> 0} (1 / (4x) ) = 0.
Note that the x axis is where y = 0, and 1 / (4 x) is the approximate y value when x is large.
Thus as x gets large, the graph approaches the x axis as an asymptote, getting closer and closer to the axis but never reaching it.
Plot a graph of y = 1 / (4x) for x values .1, 1, and 10. This will show you why 1 / (4 x) has an asymptote at the x axis.
I just don't understand the terminology they use.
Again in another problem:
R(x) = 8x^2 - x + 2 / 4x^2 - 1 I know that since the degree of the numberaor (2) is equal to the degree of the denominator (2) it is improper and we use log division.
Doing long division in itself is difficult for me to understand. But at the end of the problem, again it says as x approaches - infinity or infinity you get
-x + 4 / 4x^2 - 1 is approx = to -x / 4x^2 = -1 / 4x and then the arrow points to 0
Then we conclude that y = 2 is a horrizontal asymptote of the graph.
The long division gives you quotient
2 - (x + 4) / (4 x^2 - 1).
The parentheses are very important here; without them the order of operations would dictate a very different result.
As x gets large, the fraction (x + 4) / (4 x^2 - 1) approaches zero for the same reasons given in my answer to the preceding question.
This still leaves the 2, which doesn't depend on x and therefore remains 2. So the limiting value of the function, as x gets large, is 2.
These examples are page 349 and 350 if you need to look at them.
I can tell from your questions what the problems are asking. In other words, you've done a very good job of posing your questions.
My next question is how do you find the domain of a rational function when it is not easy to spot? Finding the domain of 4x / x - 3 is easy to spot, but when it is a problem like 5.2.16 Q(x) = -x(1-x) / 3x^2 + 5x - 2
Is there a way to solve this by using algebra? I tried to do it this way but was more confused:
3x^2 + 5x - 2 = 0
3x^2 + 5x = 2
3x^2 = 2 or 5x = 2
You need to recognize that 2 x^2 + 5 x - 2 is a quadratic equation. To solve it you either need to factor or use the quadratic formula.
The equation factors as (3 x - 1) ( x + 2) = 0, giving you 3x - 1 = 0 or x + 2 = 0. Your solutions are x = 1/3 and x = -2.
Both of these x values make the denominator zero, and both must be excluded from the domain.
Here is where I got a little confused about this. If you have 3x^2 = 2 would it become
x^2 = 2/3
x = +- sqrt(2/3)
sometimes I think my brain doesn't process this before I go on to another chapter and I forget everything I have learned up to that point.
Did I approach that one completely wrong?
Not completely; you had the right idea. Most students don't remember to look for quadratic equations until they're run into them several times after completing the chapter. Your solution had the right idea (so it's clear you've retained the overall idea of the quadratic, which is good), but instead of factoring you used a grouping pattern.
I have only reached section 5.3 because it has taken me so long to process what I have read. I am really trying to get through this chapter so I can move on to the next. Do you think at this point I would be able to take the chapter 4 test and comprehend it?
If you've gotten through section 5.3, you should be ready for the test.