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.

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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.

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look at the given polynomial f(x) whats the degree of f(x) ?
im not given any additional information.. just was given the problem and asked to find that information..
Ahh, actually im asking you a question to see if you know how to find the "degree"

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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?
oh!! i thought would be the exponent
it is indeed a polynomial, but do not multiply yet, i think we can finish the problem with out multiplying
yes, whats the degree of f(x) ?
i mean the degree is the exponent*
2
exponent of what
but then again I'm thinking its just 1
x^2?
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
for part a, the power function that the graph of f resembles for large values of |x| is \(\large -x^4\)
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?
ok...
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 \(-\color{Red}{x}^{2+1+1} = -\color{red}{x}^4\)
what does finding the degree do for me though?
good question, when you plugin large values of |x|, the function behaves almost same as the power function\(\large -x^4\)
I've been working similar problems and haven't needed that information to find the x and y intercepts
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|.`
wow... ok
you can see here, how closely the graph of f(x) matches with the power function \(-x^4\) https://www.desmos.com/calculator/vj2xhutn6z
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.
ok yes!! i see that now.
ok so finding the degree will give me the power function..
Sorry lot of questions... i think I'm just having a hard time understanding what I'm being asked to find.
good, lets find the x intercepts
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..
but now I'm not seeing how that would work.
thats right! to find x intercepts, you want to solve \[\large -x^2(x-1)(x+3)=0\]
is this problem similar to the problem f(x)={3+x If x<0 // x^2 if x> not equal to 0....
use zero product property to get : \(-x^2=0\) or \(x-1= 0\) or \(x+3=0\)
that gives you x intercepts : \(0, 1, -3\)
awesome!! ok so thats what i was thinking..
then id have to find the y intercept?
yes, simply put \(x=0\) in f(x) to get the y intercepts
f(x)=-x^2(x-1)(x+3) f(0) = ?
-3
wouldn't i substitue all x values with 0?
yes, but -3 is wrong
f(x)=-x^2(x-1)(x+3) f(0) = -0^2(0-1)(0+3) = ?
+3
no, try again
what do you get when u multiply something by 0 ?
0
so..
so the answer is 0
cuz im retarded and looked at the problem as if the 0 in front of the parentheses was invisible lol...
haha happens :)
so the y intercept is 0
how do we answer the last part `Determine whether the graph crosses or touches the x-axis at each x-intercept.`
plot the points and read the graph and how it falls
look at the graph https://www.desmos.com/calculator/bdlowvla6f
notice that the graph is crossing x axis at x=-3 and x=1 but it is just touching x axis at x=0
ok so it touches the x axis at 0,1,-3 but only crosses at 1 and -3
ooh you answered before i could finish typing.. but that is what i saw
very good!
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
the graph goes forever on both ends, it sinks down : |dw:1440390580159:dw|
zoomout in that desmos link and see...
but what about the part that is between (0,0) and (1,0)?
what about it ?
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.
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
|dw:1440390931586:dw|
if you see, the graph is just touching the x axis at x=0, it is NOT crossing the x axis
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?
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.
the turning points happen where the function has 0 slope
|dw:1441149899004:dw|
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
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
you put a point where u calculating the turning point for example(0,0) is a turning point
|dw:1441150082317:dw|
is there a way to send you what I've written down and graphed?
take a picture and attach file on here
or click on draw and roughly draw what u have
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