How to graph a rational function?

Question

I am trying to graph the function

$$y (x) := \frac{x+4}{x-2}$$

but I don't understand how to derive points and roots from the function to use for my graph. How to find these points?

Answer 1

As $x\to \pm\infty$ we have $y\to 1$. That gives a horizontal asymptote.

There is a vertical asymptote at $x=2$. Coming from the left ($x\to2^-$), we get $y\to-\infty $. Coming from the other side ($x\to2^+$), $y\to \infty $.

You can kind of fill in the rest.

Answer 2

first you should determine the domain of the function which means specifying the undefined point(s), here the domain is $ℝ-\left\{2\right\}$, if your input is $2$ then the function would be undefined, then take the derivative in order to determine the global and local extrema,hence we have:

$$\frac{dy}{dx}=\frac{\left(x-2\right)-\left(x+4\right)}{\left(x-2\right)^2}$$

now find the roots of the derivative:

$$\frac{dy}{dx}=\frac{\left(x-2\right)-\left(x+4\right)}{\left(x-2\right)^2}=0$$

$$\left(x-2\right)-\left(x+4\right)$$ clearly the derivative function does not have any real root.

now in order to determine the inflection point(s), take the second derivative:

$$\frac{d^2x}{dx^2}=\frac{12\left(x-2\right)}{\left(x-2\right)^4}=\frac{12}{\left(x-2\right)^3}$$

now find the roots of the second derivative:

the second derivative does not have any real root, means the function does not have any inflection point.

now in order to determine horizontal asymptote(s) you should calculate $\lim_{x\rightarrow +∞}f(x)$ and $\lim_{x\rightarrow -∞}f(x)$. hence we have:

$$\lim_{x\rightarrow +∞}\frac{x+4}{x-2}=1$$

and

$$\lim_{x\rightarrow -∞}\frac{x+4}{x-2}=1$$

so $y=1$ is the horizontal asymptote of the function.

now it's the time to determine vertical asymptote(s) which is $x=2$,then calculate $\lim_ {x\rightarrow 2^+}\frac{x+4}{x-2}$ and $\lim_ {x\rightarrow 2^-}\frac{x+4}{x-2}$, we have:

$$\lim_{x\rightarrow 2^+}\frac{x+4}{x-2}=+∞$$

and

$$\lim_{x\rightarrow 2^+}\frac{x+4}{x-2}=-∞$$

finally find the $x$-intercept (root(s) of the function) and $y$-intercept which are respectively $(-4,0)$ and $(0,-2)$

now you can graph the function.

Answer 3

To graph this function (an equilateral hyperbola), I have to determine the asymptoties, the centre and the simmetrical axis. I can easily find the asymptoties by doing $y=\frac{x+4}{x-2}=\frac{x+-2+6}{x-2}=1+\frac{6}{x-2}$. So the orizontal asympotote is $y=1$ and the vertical asymptote is $x=2$. The centre $C$ of the equilateral hyperbola has coords: $(2,1)$ and the simmetrical axis have equation: $y=x-1$ and $y=-x+3$.

Here the graph: https://www.desmos.com/calculator/9dqwwruqjs