Sketching graphs : Most importaint points I'm currently studying for a test which places a lot of emphasis on sketching graphs of certain functions, without anything but a ruler and a pencil. I mean tricky functions, for example:


*

*$y =  \sin^2(x)$

*$y=\dfrac{1-x-x^2}{x^2}$
I always try to sketch these functions by rewriting them to a form I can immediately jot down, or simplifying them in some way or another. For example, for $y=\sin^2(x)$ I thought I'd just take the absolute of $y=\sin(x)$.
But it's still tough for me to actually accurately sketch functions. An example of a problem I have is that for functions such as the second, I have trouble figuring out whether a sketch are 2 non-continuous lines or 1 continuous one..
So my question is:


*

*What is a good approach to these sketching questions? Is there a certain sequence which I can follow in order to make it easier (for example, always look at $x=0$ first, then find asymptotes, then...)?

 A: Some key points:


*

*Zeroes

*Local maxima/minima

*Boundary/ies of the domain, if the domain is bounded from at least one direction

*Boundary/ies of the support (where the function is non-zero), if the support is bounded in at least one direction

*Asymptotes, if such exist

*Points of inflection
Application:
$$y=\frac{1-x-x^2}{x^2}.$$
First, let's look for zeroes: $1-x-x^2=0$ when
$$x=\frac{-1\pm \sqrt 5}2.$$
So estimate $\sqrt 5$ and use that, perhaps noting the exact values.
Next, we can rewrite this as
$$y=\frac{1-x}{x^2}-1.$$
As $x$ increases or decreases without bound, this whole thing approaches $y=-1$, so draw that line.
$$y'=\frac{-x^2+2x(x-1)}{x^4}=\frac{x-2}{x^3},$$ which is $0$ exactly at $2$.
$$y''=\frac{x^3-3x^2(x-2)}{x^6}=\frac{-2x+6}{x^4},$$ which is positive at $2$, so $x=2$ is a local minimum. Draw that.
The function is undefined at $x=0$ and in fact has no limit. Figure out which ways it goes.
A: User dfeuer has given some very good hand-holds to get at the graph. I'd add that once you get all the $x$-coordinates of local maxima, minima, and inflection points, test between those points for where the graph is increasing/decreasing and concave up/down. If you know almost without thinking how the four shapes basically look (increasing+concave-up, decreasing+concave-up, increasing+concave-down, decreasing+concave-down), then you can quickly spline those shapes together, using the $(x, y)$-coordinates of local max's and min's and inflection points as anchors. Extra points for accuracy if you also get the slopes at the inflection points. It might help to organize all this using a table. 
A: I'm going to assume that you are taking calculus:
Sketching graphs is really all about translating what you know about a function unto a Cartesian plane.
Here is how I usually do it:

*

*Find the domain and range

*Find the end behavior of the function

*Find the local minima/maxima

*Find infection points

*Find zeroes

*Sketch

I will do an example here for you:
Let us consider the function $\sin ^2 (x)$, what does its graph look like?
First I will find the domain and range:
Domain: Since the $\sin^2 (x) $ function does not have any holes (and is continuous), we must check three cases: real positive numbers, real negative numbers, and zero. Since
$ \sin^2\left(\frac{\pi}{2}\right) = 1 $ we see that the function works for positive reals, since $ \sin ^2 \left(\frac{-\pi}{2}\right) = 1$ we see it works for negative reals, and the zero case is $\sin^2(0)=0$. (Note that this is not rigorous at all, you could also just rely on the domain of the sine function, and see that it is all real number)
Range: Since the $\sin (x)$ function oscillates between $\frac{+}{-} 1$  then we can see that this function does too, but since anything real to the second power is positive, this function is positive only. We will leave it at that and find the actual oscillation when we find local minima and maxima.
Next is end behavior:
We can take the limit as $x$ goes to infinity, but for this function it does not exist (the function happens to oscillate until infinity). but I think that from our knowledge of the sin function we can know that this function oscillates until infinity.
Next are the local minima and maxima:
For this we need derivatives: Remember that the slope of a tangent line at a maxima or minima is zero, so we just set the derivative equal to zero. Since
$$ \frac{d}{dx} \sin^2(x) =  \frac{d}{dx} f( g(x)) = f'(g(x)) \cdot g'(x) = 2\sin(x) \cdot \cos(x)$$
we can set that equal to zero $$ 0 = 2\sin(x)\cdot \cos(x)$$ wich will be true whenever $\sin (x)$ is equal to zero (i.e $ \pi n , n \in \mathbb{z}$) and when $\cos(x)$ is equal to zero (i.e $ \pi n + \frac{\pi}{2}, n \in \mathbb{z}$). Now we plug these values back into our original function:

*

*$\sin^2(0) = 0$

*$\sin^2\left(\frac{\pi}{2}\right) = 1$

*$\sin^2\left(-\pi + \frac{\pi}{2}\right) = 1$

*$\sin^2(- \pi) = 0$
So we can see that this graph oscillates between 1 and 0 with a wavelength of $\frac{\pi}{2}$
next we should do inflection points, through second derivatives, but for this function it is unnecessary because we have enough information to graph it. So our sketch might look like:
This
