Why does the graph not show the vertical asymptote?

Question

I have a rational function in which the denominator is equal to zero when $x = -1.4142135$.

So:

$f(x)=\dfrac{1}{x + 1.4142135}$

Then the vertical asymptote at $x = -1.4142135$ is pretty clear.

But if I had the function

$g(x)=\dfrac{x^2 - 2}{x + 1.4142135}$

... then it just becomes a straight line? What is going on here? I know that if I instead of $(x + 1.4142135)$ had $(x + \sqrt{2})$, then I would be able to cancel some expressions out and find out that the function is a linear equation with a removable discontinuity.... but I don't have the $\sqrt{2}$ in the denominator, but simply an approximation, so I shouldn't be able to make it into a linear equation, right? But it is graphically?

Answer 1

Let's consider two methods for graphing a function: method #1 is to plug in numbers and plot points, for which you can plunk down a few dollars and purchase a cheap graphing calculator; method #2 is to apply algebra and calculus and training and deep understanding and intuition, for which you can pay the tuition of a few years of mathematical training. Which one is cheaper? Which one is more sophisticated? Which one is better?

Graphing calculators are not very sophisticated creatures. If you ask a graphing calculator to compute a decimal value for $\sqrt{2}$, it will only give you an approximation consisting of finitely many digits. Of course, it can only fit finitely many digits on its screen. But its worse than that: a graphing calculator cannot even compute all of the infinitely many digits. And, when you come to think of it, even an IBM Megamind 20000 cannot compute all of the infinitely many digits.

So yes, you are right, $\sqrt{2}$ is only approximately equal to $1.4132135$, and the graph of the function $$y=\frac{x^2-2}{x + 1.4142135} $$ has a vertical asymptote at $x=-1.4142135$.

I would hazard to guess that this problem was constructed to detect whether the student's training had surpassed the capabilities of a cheap graphing calculator. At least the cheap graphing calculator recognizes that this graph also comes very, very, very close to the line $y=x - \sqrt{2}$ for a very, very long time.

Answer 2

Your graphing calculator sees $1.4142135$ as $\sqrt{2}$ altough it's just approximation. Keep in mind that your graphing calculator might not have enough capabilities. Even an online calculator such as Wolfram Alpha or Desmos shows a right line.

You are right in this case. This is just another example of why you shouldn't always rely on your graphing calculator! ;-)

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Summary:

You are right and the graphing calculator is wrong.

$f(x)=\frac{x^2-2}{x+\sqrt{2}}$ is a straight line with a removable discontinuity (hole) at $x=-\sqrt{2}$.

$g(x)=\frac{x^2-2}{x+1.4142135}$ is a hyperbola with a vertical asymptote at $x=-1.4142135$.