Can the graph of a rational function have three vertical asymptotes?
Yes, you can have as many vertical asymptotes as there are zeroes in the denominator. Take for example: f(x) = 1 / [( x - 1 )( x - 2 ) ( x - 3)]A vertical asymptote is where the graph of the rational function is undefined. This happens precisely when the the value of the function is ( c / 0 ) where c is some constant (for simplicity pretend C can't be 0). So going back to the function mentioned earlier. If you substitute x = 1 , x = 2 or x = 3, the value of the function is 1 / 0 which is undefined. The graph will produce vertical asymptotes at those points. As you can imagine you can easily conceive a rational function with as many vertical asymptotes as you so desire.
Find the vertical asymptotes, if any, of the graph of the rational function. f(x)=x/x(x-2)?
it's 2. when x is a number that could make the function undefined, meaning divided by 0, there are vertical asymptotes. so you need to find the zeroes of the denominator. x(x-2) = 0 x = 0 or x-2 = 0 x = 0 or x = 2
Find the vertical asymptotes (if any) of the graph of the function?
I'm having trouble with this one. I'm getting a result of -2 but the website is saying that's inaccurate when I submit it. Can you please help me out? Find the vertical asymptotes (if any) of the graph of the function? (Use n as an arbitrary integer if necessary. If an answer does not exist, enter DNE.) h(t) = (t2 – 2t) / (t4 – 16) Thanks
How come graphs never touch vertical asymptotes but sometimes touch horizontal asymptotes in the center of the graph (not the ends)?
Good question.Graphs approach horizontal asymptotes for large values of x in the positive and negative directions.Graphs approach vertical asymptotes when the denominators are zero.For example if (x – 3) is in the denominator then when x is just under or just over x = 3, the y value is massively big either upwards or downwards.In simple terms y = 1 + “a little bit” and the little bit gets smaller as x gets bigger.However, this is only true when x is large.The value of y is not always 1 + “a little bit”.When x is small then the y value can be big!Just think about this equation again:If x = 100 then y = 1 – 0.02 obviously approaching y = 1 from below____________________________If x = 3.01 then y ≈ – 50If x = 2.99 then y ≈ + 50 (obviously a vertical asymptote at x = 3)____________________________If x = 1.01 then y ≈ – 151If x = 0.99 then y ≈ + 151 (obviously a vertical asymptote at x = 1)____________________________If x = –100 then y = 1 + 0.02 obviously approaching y = 1 from above._____________________________Now here is the really interesting bit!WHERE IS Y EQUAL TO 1 ?so 5 = 2xand x = 2½ , y = 1The graph crosses the horizontal asymptote at x = 2 ½Here is the graph:A graph can cross a Horizontal asymptote but not a Vertical one.
How can you find asymptotes on a graphing calculator?
Yes you can find them but, the procedure is a real pain to work with. I will try to post the solution if no-one else does.Horizontal asymptote... Once the function is graphed.... eg y=1/(x-1) Obviously when x=1 the function is undefined...you can see this with your calculator. Press second (yellow button) , trace, minimum [enter] now you need to select your bounds..you can either type a value for x on the LHS (left hand side) and a value for x on the RHS or do it manually.... for the example above select the first value as 0.6 and the second value as 1.2 The output will be x=1.0000004 and y= -22906200... This is what you expect.. when x = 1 the minimum is infinity... you can do the same thing for the maximum using the same steps >Press second (yellow button) , trace, maximum [enter]Also you can use similar steps to find the Horizontal asymptote.You can try this online ti 84 graphing calculator this is really easy to use and get the most sophisticated graphing calculation results.
What are the vertical asymptotes of this SEC equation?
Vertical asymptotes occur where y is something divided by 0. We cannot divide by 0 and get a real number. Instead the graph reaches towards infinity. 1) The secant is defined as follows. sec(z) = (1 / cos(z)). So every time that the cosine is 0, that's division by 0 and you have a vertical asymptote. The plain old cos(x) function is 0 at π/2 and 3π/2 (90 degrees and 270 degrees) and, in general, at any odd number times π/2. Those are written as (2k+1)*(π/2) for any integer k because the 2k is always even so 2k+1 is always odd. This goes for all integers k, negative, 0, and positive. Now the (1/2) isn't going to affect this but the (x-π/6) will. The normal cos(x) = 1 at x = 0. But you're subtracting π/6 from it, so x has to be π/6 for the cosine to be 1. You've shifted the graph to the right by π/6. I have drawn you a graph. The cos(x-π/6) is in red, sec(x-π/6) is in black and the asymptote lines are in green. To see this graph click on the following link: http://i369.photobucket.com/albums/oo133/gerryrains/Secant_Asymptotes.jpg Notice how the cosine and secant graphs seem to sway to and from one another, meeting only where the cosine is 1 and the secant which is 1/cosine is of course also 1. I had to draw the asymptotes one at a time because they are vertical lines so I only drew enough to give you the idea. The equations are to the left. They have the curly lines in front of them. I cannot increase their size nor those of the axes' labels. The two labels with letters are labels. Ignore them. Notice that I labeled the x-axis for you in terms of pi. More information. Since sin(x) and cos(x) do periodically equal 0 the following functions all have infinitely many evenly spaced asymptotes: tan(x) = (sin(x) / cos(x)) cot(x) = (cos(x) / sin(x)) sec(x) = (1 / cos(x)) csc(x) = (1 / sin(x)) .
Why can't a graph have more than two horizontal asymptotes?
A horizontal asymptote is where the x values go off to infinity in that direction as the y values approach some value. There are only two directions it could go off to infinity - positive and negative. There's your two. For there to be more would mean the y values would approach two different numbers at the same time. This is not possible. Now, it's different for a vertical asymptote because they usually represent breaks in the domain, i.e. division by zero. You can have an infinite number of those. But unless we're talking multivariate/vector calculus, there is a maximum of two horizontal asymptotes. EDIT Jesse is wrong. in order to be a function, the range of y=arctan(x) (or the domain of x=tan(y), as he called it) must be restricted from y = (-pi/2, pi/2) because of the vertical line test, remember? There are still only two asymptotes, sorry.
How many horizontal asymptotes can the graph of y=f(x) have?
The graph of y=f(x) can have two horizontal asymptotes. One as x approaches positive infinity and one as x approaches negative infinity.
How do you find the horizontal and vertical asymptotes of a curve?
You can find the horizontal asymptotes of any function by taking the limit as x approaches infinity and negative infinity. (In the case of a demand curve, only the former should be necessary.) If the horizontal asymptotes are nice round numbers, you can easily guess them by plugging in numbers which are very large in magnitude and seeing what number the answer is close to.You can find the vertical asymptotes by checking all the places where the function is undefined. You're usually looking for divisions by zero or logarithms. Alternately, you can use a graphing utility to look for apparent vertical asymptotes. Then you take the limit of the function as it approaches the apparent location of the asymptote to check that the function actually goes to infinity or negative infinity at that location.As a difficult example, lets find the horizontal asymptotes of [math]\arctan x[/math]:[math]y = \lim_{x\to\infty} \arctan x[/math][math]\cos y = \lim_{x\to\infty} \cos (\arctan x)[/math] (this step is only valid if the limit exists, which it does, as the graph readily shows)[math]\cos y = \lim_{x\to\infty} \frac{1}{\sqrt{1+x^2}}[/math](a useful identity, easy to demonstrate by considering a right triangle with sides of 1,x, and [math]\sqrt{1+x^2}[/math])Since the denominator of the limit clearly goes to infinity, the limit clearly goes to zero. Therefore:[math]\cos y = 0[/math]and so[math]y=\frac{\pi}{2} + 2\pi k[/math]for some integer k. We can easily figure out k by plugging a very large number into the original function for x. It will be very close to [math]\frac{\pi}{2}[/math] so k must be zero.For the other horizontal asymptote, the argument proceeds equivalently, except at the end, we plug in a large negative value, see that it gives an answer very close to [math]-\frac{\pi}{2}[/math], and guess that k must be -1.