I have some troubles with the general idea of curve sketching.

I have following function that I have to sketch. $$ y = \frac{1}{2x^{2} + 12x + 6}. $$

I have the intercepts $y = 1/6$ and no $x$-intercept.

The Horizontal asymptote of $y = 0$ and the vertical asymptote of $x = 0.55$, $-5.45$.

There is a max at $(-3,-1/12)$.

Calculating those informations are no problems for me.

However when it comes to sketch the curve I always make mistakes. I have problems by drawing the line because I don't know from which side or which direction it goes when $x = +∞$, $-∞$. How do you sketch this curve step by step.

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Thank you very much in advance.


The way you figure out whether it goes up or down in general is to pick points just next to the critical points of interest, and figure out if it is positive or negative, or a little more or a little less than its value at the critical point.

For your example, let's figure out what it looks like at the far right of the diagram. Here we have passed all the relevant points of interest, and your calculations tell you that it is going to zero. But is it approaching zero from above or below? Take the function $f(x)=\frac1{2x^2+12x+6}$ and plug in a really large value of $x$. You don't need to do this with a calculator, just pick a number that you know will be bigger than everything else in the problem, say $x=1000000$. In this case you have the denominator being $2(1000000)^2+12\cdot1000000+6$, which is quite obviously positive and large, since the first term is so big that the others are just small potatoes compared to it (although in this case they are also positive). Since the numerator is just $1$, it is dominated by this big number.

This tells you (a) that it goes to zero, and (b) that it does so from above, because the reciprocal of a large positive number is a small, positive number. Of course, for the above analysis there's no need to actually plug in $x=1000000$; it suffices to note that $2x^2$ will dominate the other terms, so it will be positive for large $x$. Similarly, at large negative $x$, $2x^2$ will be large and positive, so we expect it to approach zero from above on the right side too.

Now let's look at the two singularities. The work we've done so far is already sufficient to figure out the behavior here, since it starts on the positive side, shoots off to infinity at the left pole, comes back from negative infinity since it's a simple pole, then doesn't cross the $x$-axis (there are no roots) and turns around to negative infinity, and comes out from the positive infinity side to meet our earlier asymptote to zero from above. But let's look at this process in a little more detail.

Near the left pole, we have the denominator $2x^2+12x+6$ going through one of its two roots. Now I'll assume you've already factored this polynomial to get some ugly roots, but what matters here is that there are two distinct roots. Since it is positive at large positive and negative values, we can be sure that this quadratic polynomial goes negative between the two roots. In other words, at the left root, it goes from small and positive to small and negative, which translates in the reciprocal to large and positive to large and negative. Similarly at the right root you get large and negative to large and positive since the graph of the denominator needs to pass back from negative to positive.


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