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Question 219068: what are the roots and asymptotes of f(x)=x^2-4x-5 over x^2+2x+1
Found 2 solutions by stanbon, jsmallt9:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! what are the roots and asymptotes of f(x)=x^2-4x-5 over x^2+2x+1
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Roots?
solve x^2-4x-5 = 0
(x-5)(x+1) = 0
x = 5 or x = -1
Those are the roots.
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Asymptotes?
solve x^2 + 2x +1 = 0
(x+1)(x+1) = 0
x = -1 or x = -1
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Comment: Since numerator and denominator have the factor x+1,
x = -1 is not a root and not an asymptote value. It is a hole.
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So x = 5 is a root and there are no vertical asymptotes.
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The horizontal asymptote is y = x^2/x^2 = 1
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Cheers,
Stan H.
Answer by jsmallt9(3758) (Show Source):
You can put this solution on YOUR website! What are the roots and asymptotes of  ?
We'll start by factoring because:- Fractions are easier to work with when they are reduced. And how do we reduce fractions? Answer: You cancel factors which are common to the numerator and denominator, if any. And to find these common factors, we have to have the numerator and denominator in factored form.
- Roots of a polynomial are easier to find if the polynomial is in factored form. And we will be finding roots of the numerator and denominator because:
- The roots of the numerator are the roots of the whole function.
- The roots of the denominator are the vertical asymptotes.
So even if the fraction does not reduce, having it factored will still be helpful
We'll start by factoring the numerator and denominator (which are both fairly easy to factor):
We can see now that we do have a common factor to cancel, (x+1), leaving:
We are now ready to find roots and asymptotes of f(x).
Roots are x values which make the function zero. Since f(x) is a fraction, then the easy way to find a root is based on the understanding that the only way a fraction can be zero is if the numerator is zero. (If this is not obvious to you then you will have to solve:  .) And since the numerator is x-5 then the root is 5. (If this is not obvious to you, solve the equation x-5 = 0.) (NOTE: On the graph of f(x), the root(s) will be the x-intercepts. So f(x) has only one x-intercept at (5,0).)
Vertical asymptotes, if any, occur for x values that make the (reduced) denominator zero. Since our reduced denominator is x+1, the there is just one vertical asymptote at x = -1. (If this is not obvious, then solve x + 1 = 0.)
Horizontal asymptotes, if any, occur if the function approaches a certain value for very large values, positive or negative. To find horizontal asymptotes, divide numerator and denominator by the highest power of x which is present. (We can do this with either the original ofr the reduced f(x). Since the reduced f(x) is easier we'll use it.) The highest power of x in the reduced f(x) is simply x. (In the original f(x) the highest power is  .) Dividing the numerator and denominator of f(x) by x we get:
Now we can analyze this equation to see what happens when x becomes a very large number. If we understand fractions, we will realize that the two fractions with an x in the denominator, 5/x and 1/x, will become very small as x becomes very large. The larger x gets the smaller these fractions get. They get so small that they have little effect on the overall value of f(x). So we can ignore these fractions for very large values of x. So, ignoring 5/x and 1/x, f(x) approaches the value of 1/1 = 1 for very large values of x. So y = 1 is a horizontal asymptote.
Further analysis of  will tell us "how" f(x) approaches 1. When x is a very large positive number, f(x) will approach 1 from below because- the numerator is a tiny fraction below 1 (because of the subtraction)
- the denominator is a tiny fraction above 1 (because of the addition)
- the fraction would be a (a number slightly less than 1)/(a number slightly above 1) which, as a whole, is a number slightly less than 1
This means that on the far right of the graph, where x is large and positive, the graph will come up from below and approach y = 1 from underneath.
Using similar logic for very large negative values for x we will find that f(x) will approach 1 from above. This means that on the far left of the graph, where x is large and negative, the graph of f(x) will come down from above and approach y = 1 from above.
Here's a graph so you can see all of the above "in action". (Note that the vertical and horizontal asymptotes are not drawn by Algebra.com's graphing software.):
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