f(x)=-x^2(x-1)(x+3)
Determine the end behavior: find the power function that the graph of f resembles for large values of |x|.
Find the x-and y-intercepts of the graph.
Determine whether the graph crosses or touches the x-axis at each x-intercept.

- anonymous

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- ganeshie8

look at the given polynomial f(x)
whats the degree of f(x) ?

- anonymous

im not given any additional information.. just was given the problem and asked to find that information..

- ganeshie8

Ahh, actually im asking you a question to see if you know how to find the "degree"

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## More answers

- anonymous

I'm pretty sure i understand how to find the x and y intercepts.. should i treat this as a normal polynomial function and cross multiply?

- anonymous

oh!! i thought would be the exponent

- ganeshie8

it is indeed a polynomial, but do not multiply yet, i think we can finish the problem with out multiplying

- ganeshie8

yes, whats the degree of f(x) ?

- anonymous

i mean the degree is the exponent*

- anonymous

2

- ganeshie8

exponent of what

- anonymous

but then again I'm thinking its just 1

- anonymous

x^2?

- ganeshie8

f(x)=-x^2(x-1)(x+3)
notice that there are 3 factors here, multiplying them out gives the exponent of leading term as : 2+1+1 = 4
so the degree of f(x) is 4

- ganeshie8

for part a,
the power function that the graph of f resembles for large values of |x| is \(\large -x^4\)

- anonymous

so as if i was to cross multiply id get an exponent of 2 for the first value and 1 and 1 on the middle and last term?

- anonymous

ok...

- ganeshie8

Exactly!
\(f(x)=-\color{red}{x}^2(\color{red}{x}-1)(\color{red}{x}+3)\)
its easy to see that \(-\color{Red}{x}^2*\color{Red}{x}*\color{red}{x}\) gives you \(-\color{Red}{x}^{2+1+1} = -\color{red}{x}^4\)

- anonymous

what does finding the degree do for me though?

- ganeshie8

good question, when you plugin large values of |x|, the function behaves almost same as the power function\(\large -x^4\)

- anonymous

I've been working similar problems and haven't needed that information to find the x and y intercepts

- ganeshie8

that degree thing helps in answering part a of question :
`Determine the end behavior: find the power function that the graph of f resembles for large values of |x|.`

- anonymous

wow... ok

- ganeshie8

you can see here, how closely the graph of f(x) matches with the power function \(-x^4\)
https://www.desmos.com/calculator/vj2xhutn6z

- anonymous

i havent been shown how to find the solution to these questions except for finding the x and y intercepts, domain/range, but not the power function.

- anonymous

ok yes!! i see that now.

- anonymous

ok so finding the degree will give me the power function..

- anonymous

Sorry lot of questions... i think I'm just having a hard time understanding what I'm being asked to find.

- ganeshie8

good, lets find the x intercepts

- anonymous

ok so from what I've understood, to find the x-intercept id want to find the value of the equation by f(x)=0, make the equation equal to 0, and finding the value of x.. thats why i assumed id cross multiply..

- anonymous

but now I'm not seeing how that would work.

- ganeshie8

thats right!
to find x intercepts, you want to solve \[\large -x^2(x-1)(x+3)=0\]

- anonymous

is this problem similar to the problem f(x)={3+x If x<0 // x^2 if x> not equal to 0....

- ganeshie8

use zero product property to get :
\(-x^2=0\) or \(x-1= 0\) or \(x+3=0\)

- ganeshie8

that gives you x intercepts : \(0, 1, -3\)

- anonymous

awesome!! ok so thats what i was thinking..

- anonymous

then id have to find the y intercept?

- ganeshie8

yes, simply put \(x=0\) in f(x) to get the y intercepts

- ganeshie8

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

- anonymous

-3

- anonymous

wouldn't i substitue all x values with 0?

- ganeshie8

yes, but -3 is wrong

- ganeshie8

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

- anonymous

+3

- ganeshie8

no, try again

- ganeshie8

what do you get when u multiply something by 0 ?

- anonymous

0

- ganeshie8

so..

- anonymous

so the answer is 0

- anonymous

cuz im retarded and looked at the problem as if the 0 in front of the parentheses was invisible lol...

- ganeshie8

haha happens :)

- ganeshie8

so the y intercept is 0

- ganeshie8

how do we answer the last part
`Determine whether the graph crosses or touches the x-axis at each x-intercept.`

- anonymous

plot the points and read the graph and how it falls

- ganeshie8

look at the graph
https://www.desmos.com/calculator/bdlowvla6f

- ganeshie8

notice that the graph is crossing x axis at x=-3 and x=1
but it is just touching x axis at x=0

- anonymous

ok so it touches the x axis at 0,1,-3 but only crosses at 1 and -3

- anonymous

ooh you answered before i could finish typing.. but that is what i saw

- ganeshie8

very good!

- anonymous

but then how do know the shape of the graph just by the points.. how do we know how far up it goes and where it lies on the rest of the graph

- ganeshie8

the graph goes forever on both ends, it sinks down :
|dw:1440390580159:dw|

- ganeshie8

zoomout in that desmos link and see...

- anonymous

but what about the part that is between (0,0) and (1,0)?

- ganeshie8

what about it ?

- anonymous

never mind i think I'm just over analyzing what i know and the "whys" to everything.. thank you for your help... i was on the right track for some of it.

- ganeshie8

its all good :) here is a question for you :
whats the reason for the graph of f(x) to just touch the x axis at x=0

- ganeshie8

|dw:1440390931586:dw|

- ganeshie8

if you see, the graph is just touching the x axis at x=0,
it is NOT crossing the x axis

- anonymous

Knowing, the x-and y-intercepts, the end behavior, the power function and where the graph is crossing and touching the x-axis at each x-intercept... how should all of that be applied to graphing the equation?

- anonymous

I'm supposed to label all intercepts and turning points... I understand that ill plot the (0,0), (1,0), (-3,0) and label those as my x-intercepts.. but i confused about how to label the turning points and how the power function is used.

- dan815

the turning points happen where the function has 0 slope

- dan815

|dw:1441149899004:dw|

- dan815

f(x)=-x^2(x-1)(x+3)
you have to solve the equation
d/dx(f(x)) =0
find d/dx(f(x)) and set to 0 and find at what x that happens

- anonymous

i know where the turning points occur.. I've already found the answers to everything.. the only thing I'm unsure about is how exactly is a turning point to be labeled on a graph and if and how a power function is to be labeled

- dan815

you put a point where u calculating the turning point for example(0,0) is a turning point

- dan815

|dw:1441150082317:dw|

- anonymous

is there a way to send you what I've written down and graphed?

- dan815

take a picture and attach file on here

- dan815

or click on draw and roughly draw what u have

- anonymous

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