Graph the function by hand & label key points on graph:

The domain is easy:

D: {x∈R}

The range:

R: {y∈R}

Next, I'd recommend you factor out an x:

f(x)=-x^5+11x^4-43x^3+69x^2-36x
f(x)=-x(x^4-11x^3+43x^2-69x+36)

now set f(x) = 0, you will get one of the zeros:

0=-x(x^4-11x^3+43x^2-69x+36)

0=-x
0=x

Then, you can factor x^4-11x^3+43x^2-69x+36 to get the rest of the zeros. You can use this video to help you with that.



The first turning point will occur after it crosses x=0, see the graph below:



Here's another video link on sketching a 5th degree polynomial:

Hi thanks for the help @bolbol.
But just a few things to point out from your video on how to find the roots of a 4th degree polynomial. From your video, 4:05, do you still have to find the quotient if you have more than one of the x’s? in my case I have 3 ( x = 1, 3, 4) For now I decided to just use one just like from the video. I used specifically 1, and the equation I represented was (x - 1)(x^3 - 10x^2 + 33x - 36). 5:26 of the video, used the remainder theorem again ( x = 3, 4).  Did synthetic division with 3], final equation, (x - 1)(x - 3) (x^3 - 9x^2 + 24x - 12) therefore wasn’t able to use the quadratic formula, but used a online calculator type solver to get my imaginary roots, and got 4.177651 + 1.077304i, 4.177651 - 1.077304i, 0.645. So I just wanted to know about that because my imaginary roots were incorrect.

hi chibi

I continued where bolbol left off, but then ran out of paper 😄 this should help you out



if you still can't finish it, i'll show you the rest

PS: i'm the author of that video!

Hello @bio_man, appreciate the help and your video that you’ve provided. Based from where you left off, I believe that I got the rest of the solution, but may I still see the rest anyways?

Of course, here's where we left-off and the rest of the answer:



Let me know if you have any follow-ups

Hi @bio_man thanks once again. I was watching the 2nd video that @bolbol provided, graphing a quintic function because I wanted to find the 3 turning points. But the problem is that the example that you had done has a constant, whereas mine did not, I have a 36x at the end of my equation. What would I have to do instead?

I sort of see what you mean. So I noticed that @bolbol initially common factored out the x from each term, and that lead to 4th degree polynomial as a factor:

f(x)=-x^5+11x^4-43x^3+69x^2-36x
f(x)=-x(x^4-11x^3+43x^2-69x+36)

What you could also try is remembering that the leading coefficient of original function is -1. When a=-1, expect the end-behavior furthest to the right to go down.

A polynomial with a degree of 5 is an odd number. This tells us that at the beginning of the graph, it'll start at the opposite direction relative to where it end, and we see that when we plot equation:



Notice the ╚┐pattern. Coupled with the fact that it has 4 distinct zeros, and one of the zeros has a multiplicity of 2 (the (x-3)^2), that means it has to turn 4 times.

Check this out when you get a chance:

Checked out the video, thanks. But how about finding the exact points of the turning points? One of them has a point of x=3 (root). So I want to be able to find the other 3 turning points, that’s just my only problem.

Hi chibi

Just to clarify, you want information on these?

Yeah

Those are called local minimums and maximums.



Other than creating a table of values between the zeros, I haven't come across a theorem that allows us to find them. UNLESS, you could use calculus; employing some calculus techniques will do it. If you're allowed to use calculus, here's how it's done:

Okay maybe not the exact...
but how about like an estimate of the max and min values?

To estimate, you could plug some x values between the zeros into the original equation, and from there you'll get a sense of how high or low the max/min is. But like I said, this is a calculus application when you start to get into these details. Are you being expected to verify those mins and maximum coordinates?