SOLUTION: I'm having some difficulty w/ a problem that asks why this:
f(x)=(x)/(x^2 + 2x - 3) creates the graph it does. I'm asked to include the asymptotes in my explanation which looks
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Question 1039079: I'm having some difficulty w/ a problem that asks why this:
f(x)=(x)/(x^2 + 2x - 3) creates the graph it does. I'm asked to include the asymptotes in my explanation which looks like -3 and and 1. I have the graph, but I'm unsure on how to explain it.
Any help is greatly appreciated!
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I'm having some difficulty w/ a problem that asks why this:
f(x)=(x)/(x^2 + 2x - 3) creates the graph it does. I'm asked to include the asymptotes in my explanation which looks like -3 and and 1. I have the graph, but I'm unsure on how to explain it.
Any help is greatly appreciated!
------
f(x) = x/[(x+3)(x-1)]
----
Since the denominator becomes zero when x = -3 and again}}} when x = 1,
f(x) has vertical asymptotes at x = -3 and again at x = 1.
----------------------------
The graph:::
-------------
Cheers,
Stan H.
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Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Without actually looking at the graph, there is quite a lot that you can tell about the shape of the graph.
This function has a single zero at (0,0). That is because the only zero of the numerator polynomial is 0 and the value of the function at x = 0 is 0.
Vertical asymptotes occur where there are zeros of the denominator polynomial and the numerator polynomial does not share those zeros. In the case of this function, the denominator has zeros at -3 and 1 and the numerator has a zero at 0, so there are two vertical asymptotes: x = -3 and x = 1.
The zero of the function (0) and the two vertical asymptotes together divided the x-axis into four intervals: )
, )
, )
, and )
.
Note that for the first two of the above intervals, the numerator is negative. Select a value from the 1st interval; -4 is convenient. A bit of mental calculation tells us that the denominator is positive at x = -4, so all values of the function in the first interval are negative.
Since the degree of the numerator polynomial is less than the degree of the denominator polynomial, the function is asymptotic to the x-axis. Considering the results of the analysis of the sign of the function in the first interval, we can conclude that as x increases without bound, the function approaches 0 from then negative side. Also, since the function is negative in the first interval, as x approaches -3 from the left, the function goes to negative infinity.
In the second interval, the numerator is still negative, but now the denominator is negative as well. That means the function is positive and as x approaches -3 from the right the function takes off to infinity.
At x = 0, the function value is zero, so it intersects the origin.
In the third interval, the numerator changes from negative to positive, but the denominator remains negative, hence, as x approaches the next asymptote at 1 from the left, the function goes to negative infinity.
In the fourth interval, both numerator and denominator are positive, so as x approaches 1 from the right, the function takes off to infinity, but as x increases without bound, the function again approaches 0.
John
My calculator said it, I believe it, that settles it
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