0487308
  • 0487308
What is an oblique asymptote? Will medal and fan!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
0487308
  • 0487308
@zepdrix
anonymous
  • anonymous
A rational function has at most one horizontal asymptote or oblique (slant) asymptote, and possibly many vertical asymptotes. The degree of the numerator and degree of the denominator determine whether or not there are any horizontal or oblique asymptotes.
anonymous
  • anonymous
thx

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0487308
  • 0487308
Does this graph have a oblique asymptote: \frac{x^2+6-9}{x-2} @cupcakeohla
0487308
  • 0487308
\[\frac{x^2+6-9}{x-2}\]
anonymous
  • anonymous
1 sec
0487308
  • 0487308
Okay.
anonymous
  • anonymous
mabey this will help ?
anonymous
  • anonymous
Predict if any asymptotes or holes are present in the graph of each rational function. Use a graphing calculator to draw the graph and verify your prediction. In Part A, we'll graph y = x/(x2 – 9). Let's take a moment to learn about rational functions with a numerator degree less than the denominator degree. There is a horizontal asymptote at y = 0, along the x-axis. We also have vertical asymptotes at the non-permissible values of the rational function. We'll begin by finding the horizontal asymptote. Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0. Now we'll find the vertical asymptotes. Vertical asymptotes occur at the values of x that make the denominator become zero. Write the equation x2 – 9 = 0. Factor the difference of squares to get (x + 3)(x – 3) = 0. The roots of the equation are x = -3 and x = 3. We have vertical asymptotes at x = -3 and x = 3. Now graph the rational function using technology. The asymptotes we predicted exist on the graph. When typing this into your graphing calculator, make sure the denominator is in brackets. Is it an error that the graph is crossing the horizontal asymptote? No. The horizontal asymptote is just a guide for the end behaviour of the graph. It's fine if intermediate points exist on the asymptote. Contrast this with vertical asymptotes, where the graph is never allowed to cross the asymptote. In Part B, we'll graph y = (x + 2)/(x2 + 1). We'll begin by finding the horizontal asymptote. Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0. Now we'll find the vertical asymptotes. Vertical asymptotes occur at the values of x that make the denominator become zero. Write the equation x2 + 1 = 0. Subtract 1 from both sides of the equation to get x2 = -1. Square root both sides to get x = root(-1). root(-1) is undefined. There are no vertical asymptotes. Now graph the rational function using technology. The asymptotes we predicted exist on the graph. When typing this into your graphing calculator, make sure the numerator and denominator are in brackets. In Part C we'll graph y = (x + 4)/(x2 – 16). We'll begin by finding the horizontal asymptote. Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0. Now we'll find the vertical asymptotes. Vertical asymptotes occur at the values of x that make the denominator become zero. Write the equation x2 – 16 = 0. Add 16 to both sides of the equation. Square root both sides to get x = root(16). This gives us x = ±4. Based on this result, we should have vertical asymptotes at x = -4 and x = 4. Now graph the rational function using technology. Use brackets for the numerator and denominator. The graph unexpectedly crosses the vertical asymptote at x = -4. What's going on here? Factor the denominator and simplify. This gives us 1/(x – 4). There is information loss when we cancel x + 4 from the numerator and denominator. As far as the graph is concerned, there is only one non-permissible value, +4. However, we know for a fact that x = -4 is a non-permissible value! How do we indicate this on the graph? Use an open circle to indicate that when x = -4, the graph does not exist. In words, we can say: "The graph is discontinuous at x = -4", or "The graph has a hole at x = -4". In Part D, we'll graph y = (x2 – x – 2)/(x3 – x2 – 2x). We'll begin by finding the horizontal asymptote. Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0. Now we'll find the vertical asymptotes. Vertical asymptotes occur at the values of x that make the denominator become zero. Write the equation x3 – x2 – 2x = 0. Factor out the x. Now factor the trinomial. The roots of the equation are -1, 0, and 2, in ascending numerical order. There should be vertical asymptotes at x = -1, x = 0, and x = 2. Now graph the rational function using technology. Use brackets for the numerator and denominator. The asymptotes at x = -1 and x = 2 are ignored. Factor the rational expression and cancel identical factors. This gives us 1/x. As far as the graph is concerned, only zero is a non-permissible value. This is the result of information loss from canceling identical factors. However, we know for a fact that -1 and 2 are non-permissible values. Draw holes in the graph where x = -1 and x = 2 to indicate this.
zepdrix
  • zepdrix
Wow that's a lot of words 0_o
0487308
  • 0487308
Yeah… maybe a step by step and not a copy-and-paste from a textbook?
zepdrix
  • zepdrix
Oblique asymptote = diagonal or slant asymptote. Remember when you do Synthetic Division or Polynomial Long Division, you end up with a `quotient` and a `remainder`. The quotient gives you the equation of your oblique asymptote. I think. I'm a little rusty on these :) I better check before I go spouting off nonsense.
zepdrix
  • zepdrix
Are you familiar with either method? Synthetic Division? Polynomial Long Division? Preference?
anonymous
  • anonymous
this
anonymous
  • anonymous
idk sorry
zepdrix
  • zepdrix
It's fine c: we got it
0487308
  • 0487308
Yes, synthetic division. I divided them and got x + 14 (remainder 19).
zepdrix
  • zepdrix
|dw:1450037793434:dw|
zepdrix
  • zepdrix
|dw:1450037863324:dw|Hmm your division doesn't sound right.
0487308
  • 0487308
Okay, I'll show you what I did, and maybe you can tell me what I did wrong.
0487308
  • 0487308
|dw:1450037959762:dw|
zepdrix
  • zepdrix
Looks good :) Not sure where the x+14 and remainder 19 are coming from lol
0487308
  • 0487308
Oh yeah… lol. I think that maybe I typed it wrong or something… thanks got your help c;
0487308
  • 0487308
I mean for
zepdrix
  • zepdrix
So we've determined this so far:\[\large\rm \frac{x^2+6x-9}{x-2}\quad=\qquad \color{orangered}{Quotient}+\color{royalblue}{Remainder}\]\[\large\rm \frac{x^2+6x-9}{x-2}\quad=\qquad \color{orangered}{x+8}+\color{royalblue}{\frac{7}{x-2}}\]
zepdrix
  • zepdrix
The quotient is the equation of your oblique asymptote.
0487308
  • 0487308
What is the remainder?
zepdrix
  • zepdrix
And your remainder value is still being divided by x-2, since it didn't divide evenly into it.
0487308
  • 0487308
Like what is it's significance?
zepdrix
  • zepdrix
Hmm that's a good question :) I'd have to think about it.
0487308
  • 0487308
Okay. One more question: How do you solve for the vertical asymptote's x-position?
zepdrix
  • zepdrix
\[\large\rm \frac{x^2+6x-9}{Vertical~Asymptotes}\]The denominator gives us the locations of the vertical asymptotes.
zepdrix
  • zepdrix
Recall that in the land of math, you can't divide by zero. \(\large\rm \frac{stuff}{0}\) <-- this is not a number, it is undefined.
0487308
  • 0487308
Okay… if the denominator is x-1, is the x-position of the vertical asymptote is -1?
zepdrix
  • zepdrix
Asymptotes are bad guys, bad. Law breakers. We say that the denominator can not equal zero. So to find asymptotes, we're trying to find those bad guys. So we'll `let` the denominator equal zero, and then figure out which x-values make that happen. So if our denominator was x-1, we would create this equation: \(\large\rm x-1=0\) and solve for x.
0487308
  • 0487308
x=1. This means that the x-value of the vertical asymptote is 1. I'm now going to guess that the y value of a horizontal asymptote is the numerator?
zepdrix
  • zepdrix
I'll try to answer this without calculus, even though it's more intuitive and fun...
0487308
  • 0487308
Explain it with calculus, and I'll try to follow :D
zepdrix
  • zepdrix
Have you ever heard of limits? :)
zepdrix
  • zepdrix
If not, that's ok. It will probably still make sense.
0487308
  • 0487308
Yeah, their the things like \[\lim_{x \rightarrow \infty} ,\] which means that x is getting closer and closer to infinity. They are also used in discontinuities, where ther is a little place in the graph that is unidentified.
zepdrix
  • zepdrix
Horizontal asymptotes tell us what the function is doing way way way off to the `far left` and `far right` of a function. (really really big x-values). So let's compare our numerator and denominator. \(\large\rm x^2+6x-9\) This is a second degree polynomial. Parabolic. \(\large\rm x-2\) This is a first degree polynomial. Straight line. So as we look way off to the right, letting x get bigger and bigger, what happens? Who is growing faster? Numerator or denominator?
0487308
  • 0487308
Um….. I think the numerator?
zepdrix
  • zepdrix
Ya, parabola grows much much faster than a straight line :)|dw:1450039000202:dw|
0487308
  • 0487308
Okay, and the parabola basically has a point where y is so big, it's practically infinity, whereas a linear function will usually (but not always) have that.
zepdrix
  • zepdrix
Yes :) So the numerator is "winning", which means the whole fraction is getting larger and larger, so no horizontal asymptote.\[\large\rm \frac{x^2+6x-9}{x-2}\qquad\to\qquad \frac{\infty}{stuff}=\infty\] It didn't get closer and closer to some numerical value like we wanted.
0487308
  • 0487308
Okay, so the numerator is always the vertical asymptote?
zepdrix
  • zepdrix
Back to the Algebra explanation. You'll only have a horizontal asymptote when `the degree of the numerator` is less than or equal to `the degree of the denominator`.
zepdrix
  • zepdrix
Here is a quick example where a horizontal asymptote `does exist`, so you can see the difference.
0487308
  • 0487308
Okay…
0487308
  • 0487308
So even the equation \[x^-1\] has a horizontal asymptote?
zepdrix
  • zepdrix
Ummm no :) That's has no fraction business going on.
zepdrix
  • zepdrix
When the degrees are the same, your asymptote is \[\large\rm \frac{3x^2+6x-9}{1x^2-4}\qquad\to\qquad \frac{3}{1}\] The bottom and top are growing at the `same rate` because they are both parabolas. So the asymptote is just the ratio of their leading coefficients.
0487308
  • 0487308
Okay, that's pretty cool.
0487308
  • 0487308
Does the equation 4x/7, for example, have any horizontal asymptotes?
zepdrix
  • zepdrix
No, there is no x stuff in the denominator.\[\large\rm \frac{4x}{7}\qquad\to\qquad \frac{\infty}{7}=\infty\]As x gets really really big, the numerator is winning, so no asymptote, the function just explodes upward really big.
zepdrix
  • zepdrix
Here is another example: \[\large\rm \frac{4x^2-6x+9}{2x^7-4}\qquad\to\qquad \frac{stuff}{\infty}=0\] In this example, the bottom is winning, Recall that when you divide by a really big number, you're getting smaller overall. Example:\[\large\rm \frac{1}{999999}=0.000001\approx 0\]
zepdrix
  • zepdrix
So in that last example there, we have a horizontal asymptote of y=0.
0487308
  • 0487308
Okay, in 7x+1/2x-9, is the vertical asymptote 9/2, and there is no horizontal asymptote?
0487308
  • 0487308
Never mind, the horizontal asymptote is 3.5
zepdrix
  • zepdrix
The vertical asymptote is 9/2, good! Since the degree of the numerator and denominator are the same, the function is approaching the ratio of the leading coefficients. horizontal asymptote is 7/2. Ah good, yes :) Nice correction!
0487308
  • 0487308
Just submitted my exam now, and got an 100! Thanks!
zepdrix
  • zepdrix
Noiceee :U np
0487308
  • 0487308
Just out of curiosity, how old are you? You seem to know all of the math ever created! :P

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