Take the function $f(x)=8x+120$. Find a function $g(x)$ which contains the point $(0,0)$ and is asymptotic to $f(x)$.

Question

Take the function $f(x)=8x+120$. Find a function $g(x)$ which contains the point $(0,0)$ and is asymptotic to $f(x)$ in the first quadrant.

I'm not even sure what kind of function $g(x)$ would be.

My thoughts are that: as $x$ approaches infinity $f(x)=g(x)$. Also at infinity: $g'(x)=8$. It's not quadratic as quadratics do not have asymptotes. Is it some kind of rotated logarithmic function?

Answer 1

Since $8x+120$ is an asymptote we can write $g(x)=8x+120+h(x)$ where $h(x)$ tends to zero as $x\to\infty$ and $h(0)=-120$ so as to make $g(0)=0$. One such function is $h(x)=\frac{-120}{1+x^2}$, so we have $$g(x)=8x+120-\frac{120}{1+x^2}$$

Answer 2

Hint:

For any constant $C$, the function $f(x)+C\cdot e^{-x^2}$ has the same asymptote as $f$.