3
$\begingroup$

I am learning precalculus and my precalculus book gives this equation:

equation 1

for this graph:

graph

But when I enter that equation into some online graph tool like Symbolab ( https://www.symbolab.com/graphing-calculator ) I get this graph:

graph from symbolab

It seems that (many) online calculators cancel (x+1)(x-1) in numerator/denominator before drawing a graph.

So, which graph of those 2 is "correct"? Why?

P.S. My previous question was downvoted and removed as "not interesting for math community". That was very rude having in mind that I am beginner, looking for a help. Perhaps I should join some other forum for math beginners but I don't know which and where?

$\endgroup$
8
$\begingroup$

The given diagram appears to be a graph of the function $$g(x)=\frac{2x^2-1}{x^2-1}$$ Although the supplied function is $$f(x)=\frac{2(x^2-1)}{x^2-1}=2\qquad\forall x\in\mathbb{R}\setminus\{-1,1\}$$ so the second diagram is correct.

$\endgroup$
  • $\begingroup$ Yes, that makes sense. Thank you. $\endgroup$ – Milan Che Aug 18 at 20:20
0
$\begingroup$

If $x$ equals $-1$ or $1$, $\frac{2x^2-2}{x^2-1}$ does not exist. Otherwise, it is perfectly fine to divide top and bottom by $x^2-1$. That yields the second graph: $y=2$ with gaps at $x=\pm1$.

I will second Peter Foreman's guess at the function of $\frac{2x^2-1}{x^2-1}$. The first graph appears to have asymptotes at $x=\pm1$ and $y=2$. $\frac{2x^2-1}{x^2-1}=2+\frac1{x^2-1}$ can get close to $2$ without ever equaling it and is equal to $1$ at $x=0$.

$\endgroup$
  • $\begingroup$ If 𝑥 does not equal −1 or 1, the fraction does not exist Don't you mean that the fraction does not exist if x DOES equal -1 or 1? $\endgroup$ – numbermaniac Aug 19 at 5:13
  • 1
    $\begingroup$ @numbermaniac oops. Easy fix. $\endgroup$ – Mike Aug 20 at 4:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.