Show that the graph of f(x) = sqrt(9x^2 - 6x + 3) has an oblique asymptote as x tends to positive infinity, find its equation, and decide whether the curve lies above or below it.

Step-by-step solution

1. 1

   Check the behaviour at infinity first. The radicand $9x^2-6x+3$ has a positive leading coefficient and discriminant $36-108=-72<0$, so it is positive for every real $x$ and $f$ is defined on all of $\mathbb{R}$. As $x\to+\infty$ the radicand $\to+\infty$, hence $f(x)\to+\infty$: the graph escapes upward, and the only possible asymptote is a slanted line.

2. 2

   Complete the square under the radical. Factor $9$ out of the $x$-terms:

   $9x^2-6x+3=9\left(x^2-\frac{2}{3}x\right)+3=9\left(x-\frac{1}{3}\right)^2-1+3$

   and since $9\left(x-\tfrac13\right)^2=(3x-1)^2$,

   $9x^2-6x+3=(3x-1)^2+2$

   This canonical form is the key move: it exposes the linear function hiding inside the square root.

3. 3

   Guess the asymptote from the canonical form. Writing

   $f(x)=\sqrt{(3x-1)^2+2}$

   the $+2$ is negligible compared with $(3x-1)^2$ when $x$ is large, so $f(x)\approx|3x-1|=3x-1$ for $x>\tfrac13$. The candidate asymptote is therefore

   $\Delta:\ y=3x-1$

4. 4

   Prove it with the conjugate trick. Set $h(x)=f(x)-(3x-1)$. This is an $\infty-\infty$ form, so multiply and divide by the conjugate:

   $h(x)=\frac{\left[(3x-1)^2+2\right]-(3x-1)^2}{\sqrt{(3x-1)^2+2}+(3x-1)}=\frac{2}{\sqrt{(3x-1)^2+2}+(3x-1)}$

   The numerator is the constant $2$ while the denominator $\to+\infty$, so

   $\lim_{x\to+\infty}\bigl[f(x)-(3x-1)\bigr]=0$

   which is precisely the definition of $y=3x-1$ being an asymptote to $C$ at $+\infty$.

5. 5

   Read the position of the curve from the sign of $h$. For $x>\tfrac13$ both terms in the denominator are positive, so

   $h(x)=\frac{2}{\sqrt{(3x-1)^2+2}+(3x-1)}>0$

   Hence $f(x)>3x-1$: the curve lies above its asymptote and approaches it from above, never touching it.

6. 6

   Summarise, and note the other end.

   $\boxed{y=3x-1\ \text{ is the asymptote at }+\infty,\ \text{ with } C \text{ above } \Delta}$

   Numerically, $f(10)-29=0.0345$ and $f(10^4)-29999=3.33\times 10^{-5}$, shrinking like $\tfrac{1}{3x}$. By the same computation with $|3x-1|=1-3x$, the asymptote at $-\infty$ is $y=-3x+1$, again with the curve above it.