Point of symmetry in $y=\frac{9x+7}{3x+12}$

Question

I got this question to find point of symmetry in $$y=\frac{9x+7}{3x+12}$$

As it is, I am not able to see a way to do this. Plotting on google

we get that point of symmetry may be in second quadrant. Still how will we find it if we don't have graph at all?

Thanks a lot!

Answer 1

Let consider a translation of the origin in $(-4,3)$ that is

• $u=x+4\implies x=u-4$
• $y-3=v\implies y=v+3$

and we obtain

$$y=\frac{9x+7}{3x+12}\iff v+3=\frac{9(u-4)+7}{3(u-4)+12}\iff v=\frac{9u-29}{3u} -3\iff v=\frac{-29}{3u}$$

Answer 2

The point of symmetry is the point where the horizontal and vertical asymptotes intersect.

The vertical asymptote can be found by setting the denominator equal to zero. \begin{align*} 3x + 12 & = 0\\ 3x & = -12\\ x & = -4 \end{align*}

The horizontal asymptote is the value the function approaches as $x \to \infty$ or $x \to -\infty$. If we divide the numerator by the denominator, we get $$9x + 7 = 3(3x + 12) - 29 \implies \frac{9x + 7}{3x + 12} = 3 - \frac{29}{3x + 12}$$ The term $$-\frac{29}{3x + 12} \to 0$$ as $x \to \infty$ or $x \to -\infty$. Hence, the horizontal asymptote is $y = 3$.

Therefore, the point of symmetry is $(-4, 3)$.