Polynomial Graph Analysis

[Guest]

PLEASE PLEASE HELP

I dont know how to get the X-Intercepts or The Y intercepts for the following Polynomial function F(X)= (x+2) (2X^2-2X+1). i am also not sure on the ending behavior of this function/graph. i have to solve w/o using a table of values. i also got confused when my teache mentioned something about finding out if there were imaginary zeros, then using synthetic divison. i cn dothem when factored but not like this. PLEASE HELP soon[/b]

 

[ting]

For the x-intercepts, given (x+2) is a factor, then f(-2)=0

Find the roots of (2X^2-2X+1).

Since the the discriminant, b^2-4ac is less than zero, the equation has no real roots. So only one x-intercept

Use the quadratic formula or complete the square to find the roots if you need to.


For the y intercept, find y when x is 0, solve f(0).

 

[tkhunny]

You can "do them" when you are given the answers, but you've no capacity on your own? I just have trouble believing that. Relax a little. Remember what you have heard and read. Try to remember the IDEAS you are studying, not just the specific processes.

For F(X)= (x+2) (2X^2-2X+1)

y-intercept is trivial. Plug in x = 0. Works almost every time.

F(0)= (0+2) (2(0)^2-2(0)+1) = 2*1 = 2

x-intercepts, given only a quadratic, is a matter of completing the square or using the quadratic formula. You have already (x+2) ==> x = -2

Use the quadratic formula to find the other two. 2*x² - 2x + 1 = 0

x = ½(2+sqrt((-2)²-4(2)(1)))
x = ½(2-sqrt((-2)²-4(2)(1)))

Ringing any bells yet?

 

[Guest]

thank you very much this is much less complex than i thought, i just was thrown off by the imaginary numbers part. but they are not used in the graph anyways