Oblique Asymptote to a Curve

Question

For a given curve: $$C: \frac {ax^2+bx+c}{dx+e} $$ where $a,b,c,d,e$ are integers. Let $f(x)=ax^2+bx+c$ .

---

Oblique asymptote can be found by long division of numerator by denominator.Here oblique asymptote is y=$\frac {a}{d}x$+ $\frac {b}{d}$.

---

Now If I were to multiply $y$ with $dx+e$. I will get $ax^2+bx+q$ , where q$\not=$c which is close to my f(x) but this doesn't get me to my original f(x) where a constant value is different.So what needs to be done to retain exact f(x) part of a curve?Please help!!

Answer 1

For large values of $x$, write $$y=\frac {ax^2+bx+c}{dx+e}=m+n x+\epsilon$$ and cross multiply $${ax^2+bx+c}=(m+n x+\epsilon)(d x+e)=d n x^2+x (d m+e n +d \epsilon) +e( m+ \epsilon)$$

Compare the coefficients : $$a=d n\implies n=\frac a d$$ $$b=d m+e n+d\epsilon\implies m=\frac{bd-a e}{d^2}-\epsilon $$ Since $\epsilon$ is very small, then the equation of the oblique asymptote is $$y=\frac{bd-a e}{d^2}+\frac a d x$$

Answer 2

In order of appearence:

• "the oblique asymptote of $f(x)=\frac{ax^2+bx+c}{dx+e}$ can be found by computing the polynomial ling division": TRUE

• "the result of the aforementioned operation is $\frac adx+\frac bd$": FALSE; check your calculations.

• "there is some $q$ such that $\left(\frac adx+\frac bd\right)\left(dx+e\right)=ax^2+bx+q$": FALSE, for the same reason as before.

• "if $\mu x+\rho$ is the oblique asymptote of $f(x)$, it may be the case that $(\mu x+\rho)(cx+d)\ne ax^2+bx+c$" : TRUE; it is in fact consistent with the definition of oblique asymptote (at $+\infty$) being the affine function $G_{\mu,\rho}(x)=\mu x+\rho$ such that $\lim_{x\to\infty} f(x)-G_{\mu,\rho}(x)=0$. The possibility that the infinitesimal quantity $f(x)-G_{\mu,\rho}(x)$ multiplied by $(cx+d)$ may become a non-zero constant is completely within expectations.