Graphing rational equations

[The Velociraptors]

Hello,
I’ve been struggling with the following problem which should be attached below this message.

Basically, I’m not sure why the hyperbola (if that’s the right term) starts above y=0, reaching a highest range of positive infinity. I know it would make sense given that both vertical asymptotes are odd, but when I picked a point to the left of -3/2, (leftmost odd VA), which I picked as -2, the sign analysis in purple comes out to be negative. I thought this indicates that the hyperbola would be below y=0. I picked multiple points such as -2, -3, and -5, but they all produce a negative on the sign analysis.
What am I doing wrong?

 

[Otis]

I’m not sure why the [graph] starts above y=0

It doesn't. (The graph lies in Quadrants I, II and III.) Can you explain what you mean by "starts" and why you'd thought the graph is positioned entirely above the x-axis?

both vertical asymptotes are odd

Please excuse me. Are you saying that -3/2 and 0 are odd numbers?

 

[The Velociraptors]

It doesn't. (The graph lies in Quadrants I, II and III.) Can you explain what you mean by "starts" and why you'd thought the graph is positioned entirely above the x-axis?


Please excuse me. Are you saying that -3/2 and 0 are odd numbers?

By starts, I meant (I thought it would mean) the hyperbola would be entirely under the horizontal asymptote, but I know this is wrong (I just don’t know how.)

-3/2 is an odd vertical asymptote, in that it has an odd degree. In this sense, 0 would also be an odd vertical asymptote.

 

[The Velociraptors]

It doesn't. (The graph lies in Quadrants I, II and III.) Can you explain what you mean by "starts" and why you'd thought the graph is positioned entirely above the x-axis?


Please excuse me. Are you saying that -3/2 and 0 are odd numbers?

I guess my question is, how do I know that the hyperbola in quadrants 2 and 3 isn’t completely in quadrant two or three.

 

[Steven G]

You have the wrong point for the 'hole'.
You also have the wrong limits for each sided limit at x=-3/2.

 

[The Velociraptors]

You have the wrong point for the 'hole'.
You also have the wrong limits for each sided limit at x=-3/2.

I got the point of discontinuity by substituting (1/2) into my simplified expression (without the point of discontinuity), and I’m still getting the same values.

-3/2 is a vertical asymptote, so how can the limits be wrong?

 

[Steven G]

I apologize, the y value for the hole is correct. Sorry about that.
Are you saying that if you have a vertical asymptote, then no matter which infinity (+ or -) has to be right.
Show us how you got your limits and we will go from there.

 

[Steven G]

What about the two one sided limits at x=0????

 

[The Velociraptors]

I apologize, the y value for the hole is correct. Sorry about that.
Are you saying that if you have a vertical asymptote, then no matter which infinity (+ or -) has to be right.
Show us how you got your limits and we will go from there.

Oh, my limits are wrong according to a teacher. But I just don’t understand two sided limits. I only know they have to do with vertical asymptotes, but I just don’t understand which should be negative infinity and which should be positive infinity without a graph. Can you explain it?

 

[Steven G]

If x is a little less than -3/2, would the fraction be negative or positive? If it is positive, then the limit is +oo.
You just can't randomly pick + oo or -oo!!

 

[The Velociraptors]

If x is a little less than -3/2, would the fraction be negative or positive? If it is positive, then the limit is +oo.
You just can't randomly pick + oo or -oo!!

I’m sorry, what is x here?

Do you mean (2x+3)?

 

[The Velociraptors]

If x is a little less than -3/2, would the fraction be negative or positive? If it is positive, then the limit is +oo.
You just can't randomly pick + oo or -oo!!

Wait, I think I understand what you mean. Suppose I take -1/2 which absolute value is less than -3/2, and I plug it into the expression and evaluate the signs. Is that what you’re saying?

 

[Steven G]

What is this about absolute value? Also, do you really believe that | -1/2 | < -3/2?????
If x is a little less than -3/2, what would the sign of g(x) be?
If x is a little more than -3/2, what would the sign of g(x) be?

 

[JeffM]

Let’s start over. You are saying so many odd or wrong things that helping you is quite difficult.

[math]g(x) = \dfrac{6x^2 + 7x - 5}{4x^3 + 4x^2 - 3x} = \dfrac{(3x + 5)(2x - 1)}{x(2x + 3)(2x - 1)}.[/math]
It is impossible for that function to be a real number if the denominator is zero because division by zero is not a defined operation with respect to the real number system.

Thus, the domain of this function must exclude any value of x such that the denominator has a value of 0. There are THREE such values, not two. They are x = 0, x = 1/2, and x = -3/2. We say that g(x) has three discontinuities.

Now consider the function [imath]h(x) = \dfrac{3x + 5}{2x^2 + 3x} = \dfrac{3x + 5}{x(2x + 3)}.[/imath]

Being rigorous, that is a completely different function because it has only two discontinuities at 0 and (-3/2).

IT IS FALSE TO SAY [imath]g(x) = f(x)[/imath] because [imath]h(1/2) =\dfrac{13}{4}[/imath] but g(1/2) is not even defined.

What we can say is [imath]x \ne - \dfrac{3}{2}, \ 0, \ \dfrac{1}{2} \implies g(x) = h(x).[/imath]

When a discontinuity shows up on a graph as a vertical asymptote, it is a fundamental discontinuity. There is nothing to do about it. However, some discontinuities are repairable. These arise when the function has a real (finite) limit at a discontinuity. Then a slightly different definition of the function can eliminate that discontinuity. Such discontinuities are technical; if we are clever, we can get rid of them. In the case of this example, we get rid of one discontinuity by replacing g(x) by h(x).

Now you seem to have lots of other questions, but does this clarify discontinuities for you?

If you are good to here, let’s talk about horizontal asymptotes next.

 

[Otis]

I got the point of discontinuity by substituting [x=1/2] into my simplified expression

Hi VR. I wouldn't report a function output as 6.5/2 because the teacher might think I need a calculator to do grade school arithmetic.

when I picked a point to the left of -3/2 ... I picked as -2, the sign analysis [is] negative ... multiple points such as -2, -3, and -5

I'm not sure what method you were taught, but you need to consider appropriate intervals in the domain (defined below), when analyzing signs.

\(\displaystyle \frac{3x + 5}{x(2x + 3)}\)

That function is not a hyperbola. It is a ratio of two polynomials, but not all such ratios are hyperbolas. We call it a Rational function, in general.

A Rational function's zeros, together with x-values for which the function is undefined, divide the domain into sub-intervals (which we usually refer to as simply 'intervals'). Ignore any point discontinuities in the domain (unless they lie on the x-axis, of course). We check the function's sign in the intervals, by using any x-value within each interval.

With your new function above, you have four intervals to check -- three of them are to the left of zero. You can't randomly pick negative values for x. You need to pick one, negative test value in each interval left of zero.

You need test only one positive value because the fourth interval is x>0.

 

[The Velociraptors]

Let’s start over. You are saying so many odd or wrong things that helping you is quite difficult.

[math]g(x) = \dfrac{6x^2 + 7x - 5}{4x^3 + 4x^2 - 3x} = \dfrac{(3x + 5)(2x - 1)}{x(2x + 3)(2x - 1)}.[/math]
It is impossible for that function to be a real number if the denominator is zero because division by zero is not a defined operation with respect to the real number system.

Thus, the domain of this function must exclude any value of x such that the denominator has a value of 0. There are THREE such values, not two. They are x = 0, x = 1/2, and x = -3/2. We say that g(x) has three discontinuities.

Now consider the function [imath]h(x) = \dfrac{3x + 5}{2x^2 + 3x} = \dfrac{3x + 5}{x(2x + 3)}.[/imath]

Being rigorous, that is a completely different function because it has only two discontinuities at 0 and (-3/2).

IT IS FALSE TO SAY [imath]g(x) = f(x)[/imath] because [imath]h(1/2) =\dfrac{13}{4}[/imath] but g(1/2) is not even defined.

What we can say is [imath]x \ne - \dfrac{3}{2}, \ 0, \ \dfrac{1}{2} \implies g(x) = h(x).[/imath]

When a discontinuity shows up on a graph as a vertical asymptote, it is a fundamental discontinuity. There is nothing to do about it. However, some discontinuities are repairable. These arise when the function has a real (finite) limit at a discontinuity. Then a slightly different definition of the function can eliminate that discontinuity. Such discontinuities are technical; if we are clever, we can get rid of them. In the case of this example, we get rid of one discontinuity by replacing g(x) by h(x).

Now you seem to have lots of other questions, but does this clarify discontinuities for you?

If you are good to here, let’s talk about horizontal asymptotes next.

Yup, I understand this.

 

[JeffM]

Yup, I understand this.

GREAT.

I want to clarify something. Discontinuities arise with respect to rational functions only where the definition of the function will require division by zero. Other kinds of function may have discontinuities for other reasons.

Repairable discontinuities are never marked by vertical asymptotes. The key point is that, if as x approaches the discontinuity, f(x) has a finite limit, we can always eliminate the discontinuity by defining f(x) in a more careful way. Things can get a little tricky here, but if we are talking about a point in an open interval, a limit requires that there be a right-limit and a left limit and that they be equal. The reason for thinking about left and right limits should be clear when you remember that limits are defined in terms of absolute value so you must consider positive and negative cases.

OK I shall write something about horizontal asymptotes. May take a while.

 

[Otis]

Yup, I understand this.

Have you finished this exercise?

 

[The Velociraptors]

Have you finished this exercise?

Well, I know how the graph is supposed to look from the key, and I understand the limits at infinity (which can be evaluated from the horizontal asymptote), I understand what was said above about ”replacing” or completely factoring a rational function to identify and get rid of holes, but I don’t understand the limits of vertical asymptotes very well. I know that one is supposed to be positive infinity and the other negative infinity as x approaches either left or right, but I don’t understand which one should be the negative infinity or the positive infinity.

 

[The Velociraptors]

GREAT.

I want to clarify something. Discontinuities arise with respect to rational functions only where the definition of the function will require division by zero. Other kinds of function may have discontinuities for other reasons.

Repairable discontinuities are never marked by vertical asymptotes. The key point is that, if as x approaches the discontinuity, f(x) has a finite limit, we can always eliminate the discontinuity by defining f(x) in a more careful way. Things can get a little tricky here, but if we are talking about a point in an open interval, a limit requires that there be a right-limit and a left limit and that they be equal. The reason for thinking about left and right limits should be clear when you remember that limits are defined in terms of absolute value so you must consider positive and negative cases.

OK I shall write something about horizontal asymptotes. May take a while.

Ohh, okay. So a discontinuity can arise when a hole doesn’t divide out completely, but is still apparent in either numerator or denominator. Then it’s either a hole or a vertical asymptote, but in both cases, there would be division by 0, which characterizes a discontinuity. Thanks for clarifying that.

Suppose there is a factor in the denominator which does’t divide out completely in the numerator. Is this still a repairable discontinuity as it isn’t a vertical asymptote? When I say the factor doesn’t divide out completely, I mean suppose there is a factor like (x-2) in the denominator and (x-2)^2 in the numerator.