anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
look at the given polynomial f(x) whats the degree of f(x) ?
anonymous
  • anonymous
im not given any additional information.. just was given the problem and asked to find that information..
ganeshie8
  • ganeshie8
Ahh, actually im asking you a question to see if you know how to find the "degree"

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anonymous
  • 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
  • anonymous
oh!! i thought would be the exponent
ganeshie8
  • ganeshie8
it is indeed a polynomial, but do not multiply yet, i think we can finish the problem with out multiplying
ganeshie8
  • ganeshie8
yes, whats the degree of f(x) ?
anonymous
  • anonymous
i mean the degree is the exponent*
anonymous
  • anonymous
2
ganeshie8
  • ganeshie8
exponent of what
anonymous
  • anonymous
but then again I'm thinking its just 1
anonymous
  • anonymous
x^2?
ganeshie8
  • 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
  • ganeshie8
for part a, the power function that the graph of f resembles for large values of |x| is \(\large -x^4\)
anonymous
  • 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
  • anonymous
ok...
ganeshie8
  • 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
  • anonymous
what does finding the degree do for me though?
ganeshie8
  • ganeshie8
good question, when you plugin large values of |x|, the function behaves almost same as the power function\(\large -x^4\)
anonymous
  • anonymous
I've been working similar problems and haven't needed that information to find the x and y intercepts
ganeshie8
  • 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
  • anonymous
wow... ok
ganeshie8
  • 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
  • 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
  • anonymous
ok yes!! i see that now.
anonymous
  • anonymous
ok so finding the degree will give me the power function..
anonymous
  • anonymous
Sorry lot of questions... i think I'm just having a hard time understanding what I'm being asked to find.
ganeshie8
  • ganeshie8
good, lets find the x intercepts
anonymous
  • 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
  • anonymous
but now I'm not seeing how that would work.
ganeshie8
  • ganeshie8
thats right! to find x intercepts, you want to solve \[\large -x^2(x-1)(x+3)=0\]
anonymous
  • anonymous
is this problem similar to the problem f(x)={3+x If x<0 // x^2 if x> not equal to 0....
ganeshie8
  • ganeshie8
use zero product property to get : \(-x^2=0\) or \(x-1= 0\) or \(x+3=0\)
ganeshie8
  • ganeshie8
that gives you x intercepts : \(0, 1, -3\)
anonymous
  • anonymous
awesome!! ok so thats what i was thinking..
anonymous
  • anonymous
then id have to find the y intercept?
ganeshie8
  • ganeshie8
yes, simply put \(x=0\) in f(x) to get the y intercepts
ganeshie8
  • ganeshie8
f(x)=-x^2(x-1)(x+3) f(0) = ?
anonymous
  • anonymous
-3
anonymous
  • anonymous
wouldn't i substitue all x values with 0?
ganeshie8
  • ganeshie8
yes, but -3 is wrong
ganeshie8
  • ganeshie8
f(x)=-x^2(x-1)(x+3) f(0) = -0^2(0-1)(0+3) = ?
anonymous
  • anonymous
+3
ganeshie8
  • ganeshie8
no, try again
ganeshie8
  • ganeshie8
what do you get when u multiply something by 0 ?
anonymous
  • anonymous
0
ganeshie8
  • ganeshie8
so..
anonymous
  • anonymous
so the answer is 0
anonymous
  • anonymous
cuz im retarded and looked at the problem as if the 0 in front of the parentheses was invisible lol...
ganeshie8
  • ganeshie8
haha happens :)
ganeshie8
  • ganeshie8
so the y intercept is 0
ganeshie8
  • ganeshie8
how do we answer the last part `Determine whether the graph crosses or touches the x-axis at each x-intercept.`
anonymous
  • anonymous
plot the points and read the graph and how it falls
ganeshie8
  • ganeshie8
look at the graph https://www.desmos.com/calculator/bdlowvla6f
ganeshie8
  • 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
  • anonymous
ok so it touches the x axis at 0,1,-3 but only crosses at 1 and -3
anonymous
  • anonymous
ooh you answered before i could finish typing.. but that is what i saw
ganeshie8
  • ganeshie8
very good!
anonymous
  • 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
  • ganeshie8
the graph goes forever on both ends, it sinks down : |dw:1440390580159:dw|
ganeshie8
  • ganeshie8
zoomout in that desmos link and see...
anonymous
  • anonymous
but what about the part that is between (0,0) and (1,0)?
ganeshie8
  • ganeshie8
what about it ?
anonymous
  • 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
  • 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
  • ganeshie8
|dw:1440390931586:dw|
ganeshie8
  • ganeshie8
if you see, the graph is just touching the x axis at x=0, it is NOT crossing the x axis
anonymous
  • 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
  • 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
  • dan815
the turning points happen where the function has 0 slope
dan815
  • dan815
|dw:1441149899004:dw|
dan815
  • 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
  • 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
  • dan815
you put a point where u calculating the turning point for example(0,0) is a turning point
dan815
  • dan815
|dw:1441150082317:dw|
anonymous
  • anonymous
is there a way to send you what I've written down and graphed?
dan815
  • dan815
take a picture and attach file on here
dan815
  • dan815
or click on draw and roughly draw what u have
anonymous
  • anonymous
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