General pre-calculus questions

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snugs814

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Hey nice board. I am new to the math forums so I will try my best to explain my questions. Here goes.
Describing the transformations, given y=3x+1\x-3 I can tell the graph is translated 3 units right, there is a vertical asymptote at x=3 and a horizontal asymptote at y = 3. Though it should be vertically expanded by a factor of 10? It looks like the 3 is plugged back into the top of the equation to get the number to expand, I just don't understand why since the asymptote says the function doesn't work at 3?

Another,y= 5x^2\x-4, there is a higher degree on top, so no horizontal asym., a vertical asym. at x=4, though when looking at the graph is looks like the parabola is flipped and I am not sure why? I was thinking there should be a negative somewhere to make it reflect?

Lastly, for y=(x+1)(x-2)\x-3, I see the vertical asym. at x=3 and no horizontal asym. Though looking at the graph, why is it so stretched out? Is the the x^2 in the numerator?

Thank you for the help!
 
Take the last one: \(\displaystyle \frac{(x+1)(x-2)}{x-3}\)

If we expand this by dividing, we get\(\displaystyle x+2+\frac{4}{x-3}\)

This means there is an oblique asymptote. It is the line y=x+2.

If you graph it you can see what I mean. You have an oblique asymptote when the degree of the numerator is ONE GREATER than the

degree of the denominator. That is the case for this one. Because the numerator is \(\displaystyle x^{2}-x-2\)

\(\displaystyle \frac{5x^{2}}{x-4}\) has an oblique asymptote at y=5x+20.

i.e. A graph can asymptotically approach other types of curves as well. Take \(\displaystyle \frac{x^{4}-x}{x^{2}}\)

This graph approaches the parabola \(\displaystyle y=x^{2}\) as \(\displaystyle x\to {\infty}\) or \(\displaystyle x\to -\infty\).

Here is a graph of \(\displaystyle \frac{(x+1)(x-2)}{x-3}\) along with its asymptotic line \(\displaystyle y=x+2\)

As you can see, it is not a parabola. That is the vertical asymptote at x=3 making it appear that way.

The other is the graph of \(\displaystyle \frac{5x^{2}}{x-4}\) with its oblique asymptote \(\displaystyle y=5x+20\)
 
snugs814 said:
Hey nice board. I am new to the math forums so I will try my best to explain my questions. Here goes.
Describing the transformations, given y=3x+1\x-3 I can tell the graph is translated 3 units right, there is a vertical asymptote at x=3 and a horizontal asymptote at y = 3. Though it should be vertically expanded by a factor of 10? It looks like the 3 is plugged back into the top of the equation to get the number to expand, I just don't understand why since the asymptote says the function doesn't work at 3?

Another,y= 5x^2\x-4, there is a higher degree on top, so no horizontal asym., a vertical asym. at x=4, though when looking at the graph is looks like the parabola is flipped and I am not sure why? I was thinking there should be a negative somewhere to make it reflect?

Lastly, for y=(x+1)(x-2)\x-3, I see the vertical asym. at x=3 and no horizontal asym. Though looking at the graph, why is it so stretched out? Is the the x^2 in the numerator?

Thank you for the help!
Click to expand...

Since you are new, I would like to point out couple of things regarding grammar of math boards to you.

First, we use / sign (in my laptop-keyboard this key is left of the shift key) to indicate division. The symbol used by you is left alone to be used by "html tags". Also we use * to indicate multiplication.

Grouping your operations in proper order (following PEMDAS) is absolutely essesntial for correct interpretation. For example, the first epression (algebraic) in your post should be written as y = (3x+1)/(x-3). The first set of parentheses are not necessary - whereas the second set is absolutely necessarry. Without those, your problem would be interpreted as:
\(\displaystyle \ y \ = \ 3x \ + \ \frac{1}{x} \ - \ 3\)

Your first equation is:

\(\displaystyle y \ = \ \frac{3x+1}{x-3}\)

This is the way I do these problems - your instructor may have shown you different way.

substitute

w = x-3

\(\displaystyle y \ = \ \frac{3*(w+3)+1}{w}\)

\(\displaystyle y \ = \ \frac{3w+10}{w}\)

\(\displaystyle y \ = \ 3 + \frac{10}{w}\)

\(\displaystyle y \ - \ 3 = \ \frac{10}{w}\)

Now substitutute

u = y-3 and
v = w/10

then we have

\(\displaystyle y \ = \ \frac{3x+1}{x-3} \ \ \to \ \ u = \frac{1}{v}\)

so the original expression was derived from an expression like y = 1/x

where

the origin was

shifted up 3 units and shifted right 3 units and the abcissa (x-axis) was shrunk by a factor of 10.
 
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