Asked by Amy
                g(x) = x^2 - x -12 / x + 1
I know the division brings it to x -2 + -10/x + 1
but these are the steps I have to follow
1. factor the numerator and denominator and find its domain.
( I know 0 is not in domain)
2. write in lowest ermas as p(x)/ q(x) and find the real zeros of the numerator that is find the ral solutions of the equation p(x) = 0
3. find the ral zeros of the denominator that is find the real solution of q(x) = 0
4. locate horizontal or oblique asymptotes determine points at which the graph intersects these asymptotes
5. using the real zeros of the numerator and denominator divide the x axis into intervals and determine where the graph is above the x axis and where below the x axis
6. analyze the behavior of the graph near each asymptote and indicate behavior on the graph.
7. graph ( this I can do )
Help me! This woman is impossible!!!!!
            
        I know the division brings it to x -2 + -10/x + 1
but these are the steps I have to follow
1. factor the numerator and denominator and find its domain.
( I know 0 is not in domain)
2. write in lowest ermas as p(x)/ q(x) and find the real zeros of the numerator that is find the ral solutions of the equation p(x) = 0
3. find the ral zeros of the denominator that is find the real solution of q(x) = 0
4. locate horizontal or oblique asymptotes determine points at which the graph intersects these asymptotes
5. using the real zeros of the numerator and denominator divide the x axis into intervals and determine where the graph is above the x axis and where below the x axis
6. analyze the behavior of the graph near each asymptote and indicate behavior on the graph.
7. graph ( this I can do )
Help me! This woman is impossible!!!!!
Answers
                    Answered by
            Steve
            
    What woman? It's just a math problem, like others in your book...
You have (x-4)(x+3)/(x+1)
domain is all reals except where denominator is zero. That is all reals except x = -1
zeros at x = 4 and -3 because that's where the numerator is zero and the denominator is not zero
no horizontal asymptotes, since the numerator has higher degree than the denominator. As x gets large, the fraction is just x^2/x = x
So, oblique asymptote is y=x
The graph intersects it at (-6,-6)
A little interval chart will show that
y<0 for x in (-oo,-3)U(-1,4)
y>0 for x in (-3,-1)U(4,oo)
At the end you say you can graph it. Well! In that case, what's the problem? The graph shows all the answers to the questions.
    
You have (x-4)(x+3)/(x+1)
domain is all reals except where denominator is zero. That is all reals except x = -1
zeros at x = 4 and -3 because that's where the numerator is zero and the denominator is not zero
no horizontal asymptotes, since the numerator has higher degree than the denominator. As x gets large, the fraction is just x^2/x = x
So, oblique asymptote is y=x
The graph intersects it at (-6,-6)
A little interval chart will show that
y<0 for x in (-oo,-3)U(-1,4)
y>0 for x in (-3,-1)U(4,oo)
At the end you say you can graph it. Well! In that case, what's the problem? The graph shows all the answers to the questions.
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