How to find equation of a graph using vertical asymptote?
Based on the graph, I need to find the equation. What I know: vertical asymptote x = 4, and opening at x = -4. I am struggling to find the rational function of the graph. y = 1/-x+4 is what I have currently, but I don`t know how to include the opening to the equation. What can I do?
$\endgroup$ 22 Answers
$\begingroup$$\displaystyle f(x)= \frac{1}{x}$ look similar to this. We can start from here.
The graph moved $4$ units to the right and $1$ unit up.
So, $\displaystyle f(x)= \frac{1}{x-4} + 1 = \frac{x-3}{x-4}$ is almost the same as your graph.
Finally, either you can say that $\displaystyle f: \mathbb R \backslash\{-4\} \to \mathbb R$ for the "opening" (you can call it removable discontinuity) and say $\displaystyle f(x)= \frac{x-3}{x-4}$
or
You can say $\displaystyle f(x)=\left( \frac{x-3}{x-4}\right)\cdot \left( \frac{x+4}{x+4} \right)= \frac{x^2+x-12}{x^2-16}$ and then it will be "obvious" that the function is not defined on $x=-4$ and thus has a removable discontinuity.
$\endgroup$ $\begingroup$I would say $$f(x) = \frac{x+4}{x+4} - \frac{1}{x-4}$$ for $x\neq \pm 4$ is more accurate than the other answer. This gives the appropriate values $f(0)=1\frac14$ and $f(5)=0$, and the function is increasing everywhere.
Note that the first term is intentionally unreduced so that its value is $1$ everywhere except at $x=-4$ where it is undefined (thus producing the "opening").
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