To find the turning point, differentiate the function and set it equal to 0...
Hi!
Here is the question:
Graph the following rational function about the asymptotes- (x-2)/(x^{3}-x^{2}-x+1)
So I factored the denominator using the factor theorem, and found that x=1 is a factor of the cubic function. I then used long division to end up with a quadratic- (x^{2}-1) , then I factored this difference of squares to (x+1)(x-1)
From what I can tell from this information is that the denominator is equal to (x-1)^{2}(x+1) , which tells me that there are vertical asymptotes at x= 1, -1
My only problem is that when I checked my answer with a graphing calculator, it shows an odd curve in-between these asymptotes with a vertex at x= -0.5
Can anyone offer help in terms of graphing this rational function? I don't know how to mathematically conclude that the vertex of the curve inbetween the asymptotes is indeed at x= -0.5
Why don't you use the internet? Look at this.
Thanks all for the replies, but I cannot take the derivative because this is advanced functions, not calculus. Also- my question isn't WHAT the graph looks like, I know what it looks like, my question is how to use PRE-calc to graph this function?
first off your graph is upside down. Check what you entered into your software.
Second the local maxima between the asymptotes is actually at x=-0.186141
and I really doubt you were expected to be able to determine that w/o calculus.
Just make sure you get the signs on both sides of the asymptotes correct.
Thanks everyone, I really appreciate the replies! I am just going to answer the question by plotting 'test points' to prove my understanding of the graph, because like you guys said, without calculus this graph will never be accurate.