Statistics How To

Vertical Asymptotes on the TI89: Easy Steps

TI 89 Calculus > How to Find Vertical Asymptotes on the TI89

In calculus, you will be asked to find the vertical and horizontal asymptotes of a function. The vertical asymptote occurs when the denominator of a rational function is zero. You don’t have to figure out zeros for vertical asymptotes by hand: these steps show you how to find vertical asymptotes on the TI89 in seconds.
how to find vertical asymptotes on the ti89
Image: GMU.EDU

How to Find Vertical Asymptotes on the TI89: Steps

Sample problem: Find the vertical asymptotes on the TI89 for the following equation: f(x) = (x – 6) / (x2 – 8x + 12)

Note: Make sure you are on the home screen. If you aren’t on the home screen, press the Home button.

Step 1: F2 and then press 4 to select the “zeros” command.

Step 2: Press (x-6)/(x^2-8x+12),x to enter the function.

Step 3: Press ) to close the right parenthesis.

Step 4: Press Enter.

Step 5: Look at the results. The resulting zeros for this rational function will appear as a notation like: (2,6) This means that there is either a vertical asymptote or a hole at x=2 and x=6.

Step 5: Plug the values from Step 5 into the calculator to mark the difference between a vertical asymptote and a hole. The numerator is x-6, so press 2, -, -4 and then press Enter to get 6. This means that f(2) = 6, confirming there is a vertical asymptote at x=-4. When x=0, the numerator is equal to -6. This confirms that there is a hole in the graph at x=-6. If the numerator is ever equal to zero, this means that there is a hole in the graph and not a vertical asymptote.

That’s How to Find Vertical Asymptotes on the TI89!

You can double check your answer with this calculator by Symbolab.

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Vertical Asymptotes on the TI89: Easy Steps was last modified: October 15th, 2017 by Stephanie Glen

6 thoughts on “Vertical Asymptotes on the TI89: Easy Steps

  1. Bernie

    BTW, you can also use logic to know how the graph is shaped too. First of all, as x goes irelcasingny large, the number part becomes irrelevant, so:(x+2)/(x-5)->x/xNow, if it goes to negative infinity, then:-x/-x=1Also, at x=0, f(0)=-2/5, which means that from x=negative infinity, where f(x) approaches 1, it goes downwards, and since the asymptote is at x=5, f(5) approaches negative infinity as you apporach from the left. At positive infinity, x/x=1. And when it approaches x=5, it has to be positive because any number above 5 will be:(above 5+2)/(above 5-5)=positive number/positive numberAnd so from the right, the curve goes up and f(5) approaches positive infinity. And so the graph will look like two separate U curves curves, one from the right at x=1 and slowly going up until it gets close to x=5, which then quickly goes up. The other is from the left at x=1, slowly going downwards, intersecting x=-2 and y=-2/5, and then when it gets close to x=5, quickly going downward.

  2. Brad Sinsar

    x, ^, 2, –, 8, x, +, 1,2,x,,,x

    Do you seriously expect us to understand this

    Wtf does ,,,x mean??????????????????????????????????????

  3. Andale Post author

    Whoops…part of the code on the page was corrupted, making the commas confusing. Sorry about that. It’s fixed.

  4. Kim

    You’re incorrect. You went from having the numerator as (x2 – 8x + 12) in the original function to entering (x2 – 8x + 12x) on the calculator. The zeros would be 2, 6.

  5. Andale Post author

    You’re right. Thanks for spotting that. Not sure how that happened, but it’s now fixed.