Graphing Function f(x)= (x-1)^3(x+1)^2(x+3): X and Y Intercepts & Helpful Tips

[1irishman]

Homework Statement

Graph the function f(x)= (x-1)^3(x+1)^2(x+3) and show any x and y intercepts
I don't understand how to determine the y values to correctly draw the graph. Help?

Homework Equations

The Attempt at a Solution

x=1 is a zero 3 times
x=-1 is a zero 2 times
x=-3 is a zero 1 time
I got a y-intercept of (0,2)

 

[Mark44]

To get the y values, just substitute an x value into the function's formula. The y-intercept is f(0), which BTW isn't 2.

f(0) = (0 - 1)^3 * (0 + 1)^2 * (0 + 3) = ?

Overall (i.e., for large x or for very negative x), the graph looks a lot like y = x^6, which looks similar to the graph of y = x^2. Because of the different zeroes, though, the graph will wiggle around some for values of x that are closer to zero.

The multiplicity of the zeroes gives you some useful information.

Near x = 1, the graph looks like y = ax^3. The sign of a can be determined by the other factors. So f(x) is roughly 4*4*(x - 1)^3 for x close to 1.
You can do the same for the other two factors to determine whether the graph crosses the x-axis at the zero or just dips down (or up) there.

Hope that helps.

 

[eumyang]

Also, just by looking at the function you can guess whether the leading term will be positive and negative. This, combined with the degree of the polynomial (degree of 6 as Mark44 said) will help you determine the end behavior of the polynomial.


69

 

[Char. Limit]

This is why derivatives are useful. The function, as well as its first and second derivative, tell you all you need to know. Of course, getting your equation into an easily differentiable form looks tedious...

 

[vela]

This is why derivatives are useful. The function, as well as its first and second derivative, tell you all you need to know. Of course, getting your equation into an easily differentiable form looks tedious...

I'm guessing the OP doesn't know about derivatives as he posted this question in the precalculus forum.

In this case, just analyzing the end behavior and the multiplicities of the zeros is enough to get you the general shape of the graph.