The equations of the asymptotes

Question

f(x)=P-  3x/(x^2-q^2)

 

The equations of the asymptotes are x = 1 , x = −1 , y = 2.

a)Write down the value of
(i) p ;
(ii) q .

 

(b) Let R be the region bounded by the graph of f , the x-axis, and the y-axis.
(i) Find the negative x-intercept of f

 

 

Answer 1

Looking at the function, you can see that we have a constant - a quotient involving x.

We can see that the order (the highest power) of the numerator of the quotient is one , while the order of the denominator is two.  Therefore as x gets large, this entire quotient gets small. That's the case anytime the order of the denominator is greater than the order of the numerator.

 

If the last statement made sense, you can skip this paragraph.

You can divide top and bottom of this quotient by the highest-ordered term to see this more clearly.  You would get 3/x in the numerator, and 1-[(q^2)/(x^2)] in the denominator.  As x gets larger and larger, the denominator becomes closer and closer to 1, and the numerator gets smaller and smaller. So this quotient goes to zero as x gets large.

 

We are given the horizontal asymptote, so in light of the above, we know P.

 

The x=1, x=-1 asymptotes are vertical asymptotes.  Vertical asymptotes occur at the values of x for which a quotient has zero in its denominator. Therefore there are factors in the denominator which equal zero at those values of x.

What must a and b be denominator factors in the form (x-a)(x-b) = 0 so that the expression = 0 when x = 1, and x = -1?

 

Foil that together or notice that it is a difference of squares, to find q.

 

Part b isn't a question the way you wrote it.  Do they just want you to label that? You'll need to graph this function.

Find the common denominator and multiply P by the appropriate factor over itself so that P and -3x are part of the same numerator.  Factor the new numerator to solve and use the factors to find the answer to part (b)(i)

 

Then graph the function to understand what is going on. Figure out whether the graph approaches positive or negative infinity as it gets close to the vertical asymptotes from either side by assigning values of x that are a small amount bigger or smaller than each vertical asymptote location. All you need to know is whether y gets positive large or negative large. Using all the information about the horizontal asymptote, roots of the function, how the graph approaches each vertical asymptote, you can sketch the graph. Part of this one kind of looks like tan x where the rest reminds me a lot of 1/x.

 

Please let me know if you have any troubles. We can arrange a tutoring session to go over this question with image and a discussion.

Answer 2

f(x) = p - 3x/(x²-q²)

 

To have vertical asymptotes at x = ±1, make the denominator equal to zero at those values.  So, q = 1.

 

f(x) = p - 3x/(x²-1) = [p(x²-1)-3x]/(x²-1)

 

                               = [px²-3x-p]/(x²-1)

 

Since the numerator and denominator have the same degree, there is a horizontal asymptote when y is equal to the quotient of

the leading coefficients.  So, if p = 2, then the line y = 2 is a horizontal asymptote.  

 

f(x) = [2x²-3x-2]/(x²-1)

 

To find the x-intercepts, solve the equation f(x) = 0.

 

If f(x) = 0, then 2x²-3x-2 = 0

 

                         (2x+1)(x-2) = 0

 

                          x = -1/2 or 2

 

So, -1/2 is the negative x-intercept.