Question

Please help!
Analyze the graph of the given function f as follows:
(a) Determine the end behaviors
(b) Find the domain of f. Give the answer in set builder notation
f(x)=(x-3)^2(x-1)^3(x+2)^2

Answer 1

\(\Large f(x)=(x-3)^2(x-1)^3(x+2)^2\)

End behavior means what happens to the function at the "ends" of the x-axis, that is, as x goes to infinity or x to -infinity.

That is going to depend on the SIGN of the leading coefficient, and on whether the DEGREE of the leading term is positive or negative.

Can you tell me what the leading term of f(x) is going to be?

Answer 2

\[(x-1)^3\] is the leading term right?

Answer 3

No, the leading term is what you would get if you multiplied that out. BUT you don't ACTUALLY need to multiply it out, to see what the leading term will be... just think about what the first term would be in each step.

\(\Large f(x)=(x-3)^2(x-1)^3(x+2)^2\)
None of the x's have coefficients, in the binomial factors, right? So each binomial product will just give you another factor of x in the front "slot"....

Answer 4

E.g.: if I multiply:
\((x-1)(x-3)\) I get \(x^2+........\) so \(x^2\) is the leading term.

if I multiply:
\((5x-1)(x-3)\) I get \(5x^2+........\) so \(5x^2\) is the leading term.

if I multiply:
\(x^5(3x-1)(2x-3)\) I get \(x^5\cdot 6x^2+.......=6x^7+....\) so \(6x^7\) is the leading term.

My point is, just look for the LEADING TERM that will result from the product.

Answer 5

Once you know that leading term \(\Large a\cdot x^n\)

If a>0 and n is even, then:
--- when x is a LARGE POSITIVE number, \(\Large a\cdot x^n\) will go to \(\Large +\infty\), since any power of a positive number is still positive, and a>0 won't change the sign.
--- when x is a "LARGE" NEGATIVE number, \(\Large a\cdot x^n\) will STILL go to \(\Large +\infty\), since an EVEN power of a negative number is POSITIVE, and a>0 won't change the sign.

If a>0 and n is odd, then:
--- when x is a LARGE POSITIVE number, \(\Large a\cdot x^n\) will go to \(\Large +\infty\), since any power of a positive number is still positive, and a>0 won't change the sign.
--- when x is a "LARGE" NEGATIVE number, NOW \(\Large a\cdot x^n\) will go to \(\Large -\infty\), since an ODD power of a negative number is NEGATIVE, and a>0 won't change the sign.

Now with that framework in mind, see if you can reason the end behavior possibilities if a<0.

Answer 6

(as a hint, you have the "easy" case here because you won't have a coefficient on the leading term, so in other words, a=1.... )

Answer 7

so it the end behavior of this graph be that it starts at -infinity and goes to +infinity. Is that correct?

Answer 8

Correct.... because the leading term is \(\Large x^?\)

Answer 9

Another way to think of it is, since you know that your leading coefficient =1 (there isn't a multiple in front of that first x), all you need to know is the DEGREE, and the degree is = the total number of zeros, counting multiplicity, so you can just count how many factors you have.

Here you have:
\(\Large f(x)=(x-3)^2(x-1)^3(x+2)^2\)
2 factors of (x-3)
3 factors of (x-1)
2 factors of (x+2)

So a total of 7 zeros (including multiplicity), and thus, the degree is 7.

The leading term is \(x^7\), which for "big negative" x's will go to - infinity, and for big positive x's will go to positive infinity.

Answer 10

ok i get what you were saying now. it just didnt quite click for me.

Answer 11

lousy drawing, lol, but something like this.... see how it's heading to -inf at the negative end of the x-axis, and to + inf at the positive end....