Graphing polynomials

Question

Sketch a graph of the polynomial $P(x)=(x-2)^2(x+1)^3$. You must plot and label the x and y intercepts and these should be the only points you plot.

How do I sketch the graph of a polynomial?

Answer 1

The first thing you want to do is identify the zeros. You should be able to find $2$ real values such that $P(x) = 0$. Once you have done this, you want to look at the behavior of the function between those zeros. That is, before the first zero, is your function negative or positive? After the first zero, is your function negative or positive? Ask yourself the same questions regarding the second zero. To answer these questions, simply plug in values of $x$ into $P(x)$ that are before, after, or between your two zeros (depending on what question you are trying to answer).

From this information, you know roughly how the curve dips and rises. It will also tell you whether the function crosses the $x$-axis or simply touches it at its zeros.

Finally, to finish up, look at the asymptotic behavior of your graph. What happens as $x \rightarrow -\infty$? What happens as $x \rightarrow \infty$?

All of this combined will give you enough to be able to produce a decent sketch of the graph.

Answer 2

By sketching, I am assuming you must do it by hand, not Google. So:

• Where does the graph go as $x \rightarrow $ negative infinity? What about positive infinity? This should tell you how to start.

• y-Intercept: What happens when x=0

• x-intercepts: The function is already nicely factorized. Find x for y=0.

Without caring about exactly where your local extrema are, you should find it easy from here.

Answer 3

Hint. Basically you need to know two things.

1. What is the general shape of a quintic graph?
2. What does the graph of a polynomial look like near a double or triple root?

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Edit. For this kind of question I think it is more helpful to post hints than a complete solution. However I am adding the following because IMO many of the answers posted here have completely missed the point of the question.

So, what is the point of graph sketching? IMO:

The point of graph sketching is to obtain a reasonably accurate picture of the curve with minimum work.

The "minimum work" requirement is one reason (not the only reason) why plotting lots of points is (usually) a really bad idea. I am sure that "minimum work" is also what this question means you to do, in view of the specification that you must not plot any points except the intercepts on the axes. What I suggest for the present example is this.

1. The graph is a quintic and the coefficient of $x^5$ is positive. Since we know the general shape of a quintic we can immediately say that the graph increases from $-\infty$ to $\infty$, maybe with some turning points in between.
2. The factor of $(x+1)^3$ shows that the graph goes through the $x$ axis at $x=-1$, making a "cubic wriggle" as it does so. (If you don't know what a "cubic wriggle" looks like, then you should be looking at the graph of $y=x^3$ before attempting this problem.)
3. The factor of $(x-2)^2$ shows that the graph touches the $x$ axis at $x=2$ and then stays on the same side of the axis. (If this is not clear, think of the graph of $y=x^2$.)
4. The graph cuts the $y$ axis at $y=4$. (I wouldn't actually have bothered with this, but the question asked for it.)

So we get the following graph. I post a hand-drawn sketch to try to make the point of how totally irrelevant graphing software should be for this kind of problem. The graph literally took me $15$ seconds to draw, including thinking time.

Final point: the graph clearly has a turning point somewhere between $-1$ and $2$. We could work it out if we really needed to know it, but it would add little or nothing to the general accuracy of our graph. Remember: minimum work!!!

Answer 4

You face a quintic polynomial : so, you know that the graph will go from $-\infty$ to $\infty$.

The $x$ intercepts are esay to find since the expression is already factorized; then $x=-1$ and $x=2$ correspond to $y=0$. So, you have points $(-1,0)$ and $(2,0)$.

The $y$ intercept is easy to find (just plug $x=0$ in the equation. So, you have point $(0,4)$.

You can go further using derivatives if you know them (if not, just skip the remaining): since $y=(x-2)^2(x+1)^3$, then $y'=(x-2) (x+1)^2 (5 x-4)$ and then the derivative cancels at $x=2$, $x=-1$ and $x=\frac{4}{5}$. These points will correspond to maximum and minimum of the graph. We already know the value of $y$ for the first two points; for the last, $(0.80,8.40)$. Now, if you study the sign of the derivative, you will notice that it is positive or zero from $-\infty$ to $\frac{4}{5}$, negative from $\frac{4}{5}$ to $2$ and positive. This gives you : increasing, decreasing, increasing.