# Various Problems for Precalculus Exam, Unit 3

1. Jan 11, 2009

### jacksonpeeble

1. The problem statement, all variables and given/known data
16. Determine the end behavior of the polynomial P(x)=x3(x+2)(x-3)2.

19. (No Calculator) Sketch the graph of the polynomial function P(x)=2x^3-x^2-18x+9. Where is P(x)>=0?

22. If P(x) has a local maximum at (1,5) and a local minimum at (-2, -4), then find the extrema of f(x)=P(x-2)+3.

2. Relevant equations
16. P(x)=x3(x+2)(x-3)2

19. P(x)=2x^3-x^2-18x+9

22. f(x)=P(x-2)+3

3. The attempt at a solution
16. Degree=3, Lead Coefficient=+, therefore $$y\rightarrow\infty$$ as $$x\rightarrow\infty$$ and $$y\rightarrow-\infty$$ as $$x\rightarrow-\infty$$. The answer key (which we have, so I don't just need the final answer) says I'm wrong.

19. The degree is odd, and the leading coefficient is positive, so the end behavior is $$y\rightarrow\infty$$ as $$x\rightarrow\infty$$ and $$y\rightarrow-\infty$$ as $$x\rightarrow-\infty$$. The +9 no doubt moves it up nine. However, what other information do I need to include to accurately sketch the graph, and how do I determine where P(x) is greater than or equal to zero?

22. What? Where did the f(x) come from? How does this work (utterly perplexed)?

2. Jan 11, 2009

### jacksonpeeble

I want to thank in advance all of the people that have been helping me with my exam review - don't worry; there are only two units left after this one! I really appreciate the advice and tips you've been giving. I have to complete some Anthropology work, but I'll check this topic again soon.

3. Jan 11, 2009

### jgens

I don't have much time to help, but reexamine the behavior of the first equation when x is negative, while x^3 is negative and (x+2) is negative, (x-3)^2 is?

Don't have time to walk through 19, sorry.

f(x) is just another function (who's values depend on P(x)). To find the maximum and minimum values of f(x) we need to find the max and min. values of P(x-2). Ex. P(x-2) should have a maximum value at x = 3; therefore, P(1) = 5. f(x) = P(x-2) + 3, hence, f(3) = 5 +3 = 8. Can you do something similar?

4. Jan 12, 2009

### NoMoreExams

You can easily factor this one, factor x^2 from the first 2 terms and 9 from the other 2
For this one, pick a function that has a max at that point and a min at the other one. Then see what happens when you convert it to P(x-2) + 3.

They brought in f(x) to define a new function.

For example if I had P(x) = x^2 and then said f(x) = P(x+2) then I would write P(x+2) as (x+2)^2 and that would be my new f(x).

5. Jan 12, 2009

### HallsofIvy

Staff Emeritus

16. P(x) has even order (6) so does NOT got to -infinity as x goes to -infinity.

17. Factor just as NoMoreExams suggested (which was very clever, by the way) and it is easy.

18. Generally, changes before the "main function" are horizontal changes in the graph (changes to x) and changes after the "main functions" are vertical changes in the graph (changes to y).
If f(x)= P(x-2)+ 3 then the graph of f(x) is exactly the graph of P(x) moved to the right 2 and up 3. P(x) is, of course, any function with max at (1, 5) and min at (-2, -4). Find the max and min of f(x) by shifting those points as I said.