<p>Given the function of f defined by f(x) = 2x-2/(x^2) + x - 2 </p>
<p>a) For what values of x is f(x) discontinous</p>
<p>b) At each point of discontinuity found in (a), determine whether f(x) has a limit, and if so, give the value of the limit. </p>
<p>c) Write an equation for each vertical and horizontal asymptote of the graph of f. Justify each answer</p>
<p>d) A rational function g(x) = a/b+x is such that g(x) = f(x) wherever f is defined. Find the value of a and b. </p>
<p>Can anyone please help me. I never understand what my calc teacher is teaching me and iunno if she knows what shes talking about =/. Thanks in advance.</p>
<p>A) First, factor the numerator and denominator. When you factor, you should get this [2(x-1)] / [(x+2)(x-1)]. In the denominator, you can see that x cannot equal -2 or 1 because it would make it undefined, and one of the definintions of continuity is that f(x) is defined. So, f(x) has a removable disconinuity at x=1 (because it can be cancelled out) and a nonremovable discontinuity at x=-2 (because it cannot be cancelled out).</p>
<p>B) There is a limit as x approaches 1 (it's 2/3), but not as x approaches -2.</p>
<p>C) Vertical asymptote at x=-2, because x approaches -2 from either side without bound.</p>
<p>Wait, is the problem (2x-2)/[(x^2)+x-2] or without the brackets?
Secondly, is g(x)=a/(b+x) or w/o the parentheses?
assuming it's the first case in both above,
D) a=2 and b=2 or 2/(2+x). because if you factor the original equation and get rid of the (x-1)'s you get it in a/(b+x) form. and these are the answers.</p>
<p>Thanks a lot guys, it makes a lot more sense now. However in part c, it asks about the horizontal asymptote. Since, you did not address it, I assume there is not one?</p>
<p>i think there's never a horizontal asymptote for linear/quadratic, and always a horizontal asymptote for quadratic/quadratic. but this might be wrong.</p>
<p>the most important thing to remember for AP Calc though is that YOUR CALCULATOR IS YOUR BEST FRIEND.</p>
<p>The general rule of thumb for horizontal asymptotes is:</p>
<p>If the numerator is a function that rises more slowly than the denominator, then the horizontal asymptote is y = 0.</p>
<p>If the numerator is a function that rises equally quickly compared to the denominator, then the horizontal asymptote is the quotient of the lead coefficients.</p>
<p>If the numerator is a function that rises more quickly than the denominator, then there is no horizontal asymptote. (The function actually has what's called a "slant asymptote" that can be determined using polynomial division.)</p>
<hr>
<p>So, as I read this particular problem, where you have a linear divided by a quadratic, you have a horizontal asymptote at y = 0. Note that this is consistent with the graph of 2/(x+2) that you have concluded is the answer for (D), where the function approaches 0 both as x -> infinity and as x -> -infinity.</p>
<p>That's the connection a lot of folks don't make. If you are looking at the limit as x goes to either positive or negative infinity, and you get a finite result c, then y = c is a horizontal asymptote for that function.</p>
<p>A different way of phrasing "rising more slowly/equally/faster" is:
Look at the highest power in the numerator and denominator;,</p>
<p>-if the numerator is a lower power than the denominator, then there is a horizontal asymptote at y=0</p>
<p>-if the numerator is equal in power to the denominator, look at the leading coefficients, and then there is a horizontal asymptote at y=numer/denom (coefficients)</p>
<p>-if the numerator is a higher power than the denominator, there is a slant asymptote. to find it divide the numerator by the denominator to get a linear equation which is the slant asymptote. (ignore the remainder)</p>
<p>The "highest power" idea works really well with polynomials, but the rule holds true even with non-polynomial functions.</p>
<p>For instance, the function f(x) = (sin x)/x has a horizontal asymptote at y = 0, because the function (sin x) always varies between -1 and 1, while y = x tends towards positive/negative infinity with a slope of 1. Therefore, the denominator overpowers the numerator.</p>
<p>
[quote]
-if the numerator is a higher power than the denominator, there is a slant asymptote. to find it divide the numerator by the denominator to get a linear equation which is the slant asymptote. (ignore the remainder)
[/quote]
</p>
<p>From what I remember from precalc: if the numerator is a higher power than the denominator, there is no horizontal asymptote, and there is a slant asymptote only if the degree of the numerator is exactly 1 higher than that of the denominator. I could be misremembering, though.</p>
<p>That's probably based on the definition of "slant asymptote" given to you in precalculus. A lot of folks like to think that the "slant asymptote" has to necessarily be a line.</p>
<p>In theory, if the degree of the numerator were two larger than the degree of the denominator, then the end behavior of the function more gradually approaches a particular quadratic. Whether that quadratic can accurately be called a "slant asymptote" is admittedly a subject of some debate, although the quadratic serves the same purpose as the linear case does when the degree of the numerator is one larger than the degree of the denominator.</p>