Question

How do you sketch the general shape of `f(x)=x^3-2x^2+1` using end behavior?

Answer 1

`x^3-2x^2+1 \to-\infty` as `x\to-\infty`

`x^3-2x^2+1 \to+\infty` as `x\to+\infty`

Explanation:

You need to know that the end behaviour of a polynomial depends on its degree:

• if the degree is even, both limits at `\pm\infty` will be `+\infty`
• if the degree is odd, you'll have the limit according to the direction: if `p(x)` is your polynomial, then `lim_{x\to-\infty}p(x)=-\infty` and `lim_{x\to+\infty}p(x)=+\infty`

This is easy to explain: an even degree means that you surely are the square of something: `x^2` is the square of `x`, `x^4` is the square of `x^2`, and so on. If `n` is even, `x^n` is the square of `x^{n/2}`. And since squares are always positive, the limits can only be `+\infty`.

On the other hand, you can see an odd power as an even power of `x` multiplied one more time by `x`. For example, see `x^7` as `x^6*x`.

We already observed that `x^6` tends to positive infinity in both direction, so at `-\infty` you'll have `x^6*x\to (+\infty)(-\infty)=-\infty`, while at `+\infty` you'll have `x^6*x\to (+\infty)(+\infty)=+\infty`.

The reason for which the leading term is the only relevant one is simple, too: let's analyze your case: we have

`x^3-2x^2+1 = x^3(1-2/x+1/x^3)`

So, if we factor the greatest power of `x`, all the remaining terms will tend to zero as `x` approaches infinity (in both direction), showing that (in this case) `x^3` is the only relevant term to investigate the end behaviour.