# Given the curve $y=\frac{5x}{x-3}$. Find its asymptotes, if any.

Given the curve $$y=\frac{5x}{x-3}$$. To examine its asymptotes if any.

We are only taking rectilinear asymptotes in our consideration

My solution goes like this:

We know that, a straight line $$x=a$$, parallel to $$y$$ axis can be a vertical asymptote of a branch of the curve $$y=f(x)$$, iff $$f(x)\to\infty$$ when $$x\to a+0$$, $$x\to a-0$$ or $$x\to a$$. Similarly, a straight line $$y=a$$, parallel to $$x$$ axis can be a horizontal asymptote of a branch of the curve $$x=\phi(y)$$, iff $$x\to\infty$$ when $$y\to a+0$$, $$y\to a-0$$ or $$y\to a$$. Using these lemmas, we obtain $$x=3$$ and $$y=5$$ as the rectilinear asymptotes. This is because, $$y=\frac{5x}{x-3}=f(x)$$, then as $$x\to 3$$, $$f(x)\to \infty$$. Also, since $$y=\frac{5x}{x-3}$$, thus,$$xy-3y=5x$$ or $$xy-5x=3y$$, hence $$x=\frac{3y}{y-5}=\phi(y)$$. Due to which $$\phi(y)\to \infty$$ as $$y\to 5$$.

Now, I tried checking, if there is any oblique asymptote. We know that, $$y=mx+c$$, is an oblique asymptote of $$y=f(x)$$, iff $$\exists$$ a finite $$m=\lim_{|x|\to\infty}\frac{y}{x}$$ and $$c=\lim_{|x|\to\infty} y-mx$$. Now, we have the function $$y=\frac{5x}{x-3}$$ and hence, $$\frac{y}{x}=\frac{5}{x-3}$$ and hence, $$m=\lim_{|x|\to\infty}\frac{y}{x}=\lim_{|x|\to\infty}\frac{5}{x-3}=0$$. Hence, $$c=\lim_{|x|\to\infty} y-mx=\frac{5x}{x-3}=\frac{5}{1-\frac 3x}=5$$. So, the oblique asymptote of the given curve is $$y=mx+c=5$$, which is same as the horizontal asymptote, parallel to $$x-axis$$ as shown above. Is the solution correct? If not, where is it going wrong?

• If $x=3$ is called vertical asymptote, I suppose $y=5$ will be called by many horizontal asymptote. It is just a name, but well, at least it is good to be clear that they point in orthogonal directions.
– plop
Jan 8 at 16:01
• You are confusing the definitions of horizontal and vertical. A horizontal line is parallel to or coincides with the $x$-axis; a vertical line is parallel to or coincides with the $y$-axis. Therefore, the line $x = 3$ is a vertical asymptote, while the line $y = 5$ is a horizontal asymptote. Jan 8 at 16:01
• @owl The line $y = 5$ is a horizontal asymptote. Jan 8 at 16:03
• @N.F.Taussig Yeah, its called horizontal asymptote, fixing it... Jan 8 at 16:05
• You can write the equation as $(y-5)(x-3)=15,$ which reveals the asymptotes. This generalizes to all hyperbolas. Jan 9 at 7:42

• Horizontal asymptotes occur when the numerator of a rational function has degree less than or equal to the degree of the denominator.
• Oblique asymptotes occur when the degree of the denominator of a rational function is one less than the degree of the numerator.

Note that a rational function can't have both horizontal and oblique asymptotes.

A homographic function is the quotient of two first-degree polynomial functions, through an expression in the form, with $$c\ne 0$$,

$$f \left( x \right)=\dfrac {ax+b} {cx+d}=\frac{5x+0}{x-3}$$

$$a=5, b=0, c=1, d=-3$$.

• If $$c=0$$ then $$y =\frac{a \cdot x}{d}+\frac{b}{d}$$, which is the equation of a line of angular coefficient $$a \over d$$, which intersects the $$y$$-axis at the point of ordinate $$b \over d$$.

If the mixed product between the coefficients $$a\cdot d=b\cdot c$$, then $$d=\frac{b\cdot c}{a}$$ can be substituted, and then, picking up by common factor, $$y=\frac{a(ax+b)}{c(ax+b)}$$, which simplified gives $$y=\frac{a}{c}$$, i.e., a line parallel to the $$x$$-axis representing the horizontal asymptote of the homographic function (The same result is reached by exploiting the definition of limit, i.e., $$y=\lim_{x \to +\infty} \frac{(ax+b)}{(cx+d)} =\lim_{x \to +\infty} \frac{x(a+\frac{b}{x})}{x(c+\frac{d}{x})} = \frac{a+0}{c+0} =\frac{a}{c}$$ which is the horizontal asymptote).

• If $$c \neq 0$$ and $$a \cdot d \ne b \cdot c$$, then the homographic function represents an equilateral hyperbola with asymptotes parallel to the coordinate axes. In particular, the asymptotes have equations $$y=\frac{a}{c}$$ and $$x=-\frac{d}{c}$$. The $$x=-\frac{d}{c}$$ is a vertical asymptote because must be $$cx+d\ne 0$$. In our case therefore long calculations are not there. In fact $$ad-bc=5\cdot(-3)-0\ne 0$$. Therefore the asymptotes have equation $$y=5$$ (horizontal) and $$x=-\frac{-3}{1}=3$$ (vertical).

You have correctly found the asymptotes of the function. Observe that we can express the function in the form $$f(x) = \frac{5x}{x - 3} = \frac{5x - 15 + 12}{x - 3} = \frac{5x - 15}{x - 3} + \frac{12}{x - 3} = \frac{12}{x - 3} + 5$$ which tells us that the graph of $$f$$ can be obtained by dilating the graph of $$y = 1/x$$ by a factor of $$12$$, shifting it to the right by three units, and up by $$5$$ units. Since the curve $$y = 1/x$$ has horizontal asymptote $$y = 0$$ and vertical asymptote $$x = 0$$, it follows that the function $$f$$ has horizontal asymptote $$y = 5$$ and horizontal asymptote $$x = 3$$.

A function $$f$$ has horizontal asymptote $$y = k$$ if $$\lim_{x \to \infty} f(x) = k$$ or $$\lim_{x \to -\infty} f(x) = k$$ Observe that \begin{align*} \lim_{x \to \infty} f(x) & = \lim_{x \to \infty} \frac{5x}{x - 3} = \lim_{x \to \infty} \left(\frac{12}{x - 3} + 5\right) = 0 + 5 = 5\\ \lim_{x \to -\infty} f(x) & \lim_{x \to -\infty} \frac{5x}{x - 3} = \lim_{x \to \infty} \left(\frac{12}{x - 3} + 5\right) = 0 + 5 = 5 \end{align*} which tells us that the only horizontal asymptote is $$y = 5$$.

In fact, a rational function can have at most one horizontal asymptote. This is not true for all functions. For instance, the function $$g: \mathbb{R} \to \mathbb{R}$$ defined by $$g(x) = \frac{x}{|x| + 1}$$ has two horizontal asymptotes since \begin{align*} \lim_{x \to \infty} g(x) & = \lim_{x \to \infty} \frac{x}{|x| + 1} = \lim_{x \to \infty} \frac{x}{x + 1} = \lim_{x \to \infty} \frac{1}{1 + \frac{1}{x}} = 1\\ \lim_{x \to \infty} g(x) & = \lim_{x \to \infty} \frac{x}{|x| + 1} = \lim_{x \to \infty} \frac{x}{-x + 1} = \lim_{x \to \infty} \frac{1}{-1 + \frac{1}{x}} = -1 \end{align*}

A function $$f$$ has a vertical asymptote at $$x = a$$ if at least one of the following four statements is true: \begin{align*} \lim_{x \to a^+} f(x) & = \infty & \lim_{x \to a^+} f(x) & = -\infty\\ \lim_{x \to a^-} f(x) & = \infty & \lim_{x \to a^-} f(x) & = -\infty \end{align*} In this case, \begin{align*} \lim_{x \to 3^+} f(x) & = \lim_{x \to 3^+} \frac{5x}{x - 3} = \lim_{x \to 3^+} \left(\frac{12}{x - 3} + 5\right) = \infty\\ \lim_{x \to 3^-} f(x) & = \lim_{x \to 3^-} \frac{5x}{x - 3} = \lim_{x \to 3^-} \left(\frac{12}{x - 3} + 5\right) = -\infty \end{align*} so $$x = 3$$ is a vertical asymptote of the graph.

A rational function can have a horizontal asymptote when the degree of the numerator is at most that of the denominator, which is the case here since both the numerator and denominator have degree one. A rational function can have an oblique asymptote when the degree of the numerator exceeds that of the denominator by one. A rational function can have a curvilinear asymptote when the degree of the numerator exceeds that of the denominator by more than one.