Graphing rational equations

[Otis]

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

Hi. That's why we test the sign within each sub-interval, as I explained in post #15. Also, there are components of deduction involved, when drawing the graph. I'll give an example of what I mean, later in this post. (You had done a sign test, but you seemed to have randomly picked negative test values. You need to test one value in each appropriate interval.)

The zeros and vertical asymptotes provide the intervals. In your function, there is one zero (-5/3) and two vertical asymptotes (-3/2 and 0). I'm ignoring the point discontinuity because a function doesn't change sign when "jumping" across a point discontinuity, unless the point discontinuity lies on the x-axis (which you've already determined is not the case).

Example: The values -3/2 and 0 define an interval. We pick a value within that interval to test: x=-1. We determine each factor's sign:

3(-1) + 5 = -3 + 5, so the sign of (3x + 5) is positive

2(-1) + 3 = -2 + 3, so the sign of (2x + 3) is positive

The sign of factor (x) is negative

One negative factor makes for a negative ratio. In other words, all function outputs are negative within that interval. Therefore, all points on the curve between x=-3/2 and x=0 have negative y-coordinates. That tells us the graph must be approaching negative infinity, both as x approaches -3/2 from the right and as x approaches 0 from the left.

Here's a bit of deduction. We now know within the interval (-3/2,0) that the graph rises from negative infinity as x moves away from -3/2, and we know the graph falls back toward negative infinity as x approaches 0. We also know that there is no zero within that interval. Therefore, we can deduce that the graph must turn around at some point below the x-axis between x=-3/2 and x=0. In order to graph that turnaround point, we would need to evaluate some additional test values -- to roughly determine the local maximum.

That's pretty much it. I tested one interval; you test the remaining three. Pick a test value within the intervals:
(-∞,-5/3),
(-5/3,-3/2)
(0,∞).

If you need additional help reasoning out the shape of the graph, let us know.

 

[JeffM]

This is about non-vertical asymptotes, not your previous question.

As far as I know, rational functions do not have degree. A rational function is the ratio of two polynomial functions. Polynomials do have degree.

A polynomial can be expressed as

[math]\sum_{i=0}^n a_ix^{(n-i)}, \text { where } n \text { is an integer } > 0 \text { and } a_0 \ne 0.[/math]
The degree of the polynomial is n.

So we can express a rational function (ignoring domain issues) as

[math]f(x) = \dfrac{P_n(x)}{Q_m(x)}, \text { where P and Q are polynomials of degree n and m respectively.}[/math]
Now obviously there are three cases: n = m, n > m, or n < m.

Suppose n < m.

[math]f(x) = \dfrac{P_n(x))}{Q_m(x)} = \dfrac{P_n(x)}{Q_m(x)} * \dfrac{1/x^m}{1/x^m} = \dfrac{\dfrac{P_n(x)}{x^m}}{\dfrac{Q_m(x)}{x^m}} \text { if } x \ne 0.[/math]
[math]\dfrac{P_n(x)}{x^m} = \dfrac{1}{x^m} * \sum_{i=0}^n a_ix^{(n-i)} = \sum_{i=0}^n \dfrac{a_i}{x^{(m+i-n)}}.[/math]
But n < m and i is non-negative by supposition so m + i - n is positive. As the absolute value of x becomes very large, all the terms in that sum approach zero, and thus the sum approaches zero.

[math]\dfrac{Q_m(x)}{x^m} = \dfrac{b_0x^{(m-0)}}{x^m} + \dfrac{}{x^m} * \sum_{j=1}^m b_j x^{(m-i)} = b_0 + \sum_{j=1}^m \dfrac{b_j}{x_i}.[/math]
As the absolute value of x becomes very large, all the terms involved in sigma approach zero.

[math]\therefore |x| >> 0 \implies f(x) \text { approaches } \dfrac{0}{b_0} = 0 \text { because } b_0 \ne 0 \text { by definition.}[/math]
Obviously that is not a formal proof, but it is I hope intuitive.

I’ll let you follow the same logic for the case of n = m. You should end up at

[math]\therefore |x| >> 0 \implies f(x) \text { approaches } \dfrac{a_0}{b_0} \ne 0 \text { because } a_0 \ne 0 \ne b_0 \text { by definition.}[/math]
To summarize this in English, if the degree of the polynomial in the numerator of a rational function is NOT greater than the degree of the polynomial in the denominator, the rational function approaches a horizontal asymptote when the absolute value of the variable becomes very large. That asymptote is the x-axis if the degree of the numerator is exceeded by the degree of the denominator. That asymptote is not the x-axis if the numerator and denominator have the same degree.

Is this clear?

 

[The Velociraptors]

This is about non-vertical asymptotes, not your previous question.

As far as I know, rational functions do not have degree. A rational function is the ratio of two polynomial functions. Polynomials do have degree.

A polynomial can be expressed as

[math]\sum_{i=0}^n a_ix^{(n-i)}, \text { where } n \text { is an integer } > 0 \text { and } a_0 \ne 0.[/math]
The degree of the polynomial is n.

So we can express a rational function (ignoring domain issues) as

[math]f(x) = \dfrac{P_n(x)}{Q_m(x)}, \text { where P and Q are polynomials of degree n and m respectively.}[/math]
Now obviously there are three cases: n = m, n > m, or n < m.

Suppose n < m.

[math]f(x) = \dfrac{P_n(x))}{Q_m(x)} = \dfrac{P_n(x)}{Q_m(x)} * \dfrac{1/x^m}{1/x^m} = \dfrac{\dfrac{P_n(x)}{x^m}}{\dfrac{Q_m(x)}{x^m}} \text { if } x \ne 0.[/math]
[math]\dfrac{P_n(x)}{x^m} = \dfrac{1}{x^m} * \sum_{i=0}^n a_ix^{(n-i)} = \sum_{i=0}^n \dfrac{a_i}{x^{(m+i-n)}}.[/math]
But n < m and i is non-negative by supposition so m + i - n is positive. As the absolute value of x becomes very large, all the terms in that sum approach zero, and thus the sum approaches zero.

[math]\dfrac{Q_m(x)}{x^m} = \dfrac{b_0x^{(m-0)}}{x^m} + \dfrac{}{x^m} * \sum_{j=1}^m b_j x^{(m-i)} = b_0 + \sum_{j=1}^m \dfrac{b_j}{x_i}.[/math]
As the absolute value of x becomes very large, all the terms involved in sigma approach zero.

[math]\therefore |x| >> 0 \implies f(x) \text { approaches } \dfrac{0}{b_0} = 0 \text { because } b_0 \ne 0 \text { by definition.}[/math]
Obviously that is not a formal proof, but it is I hope intuitive.

I’ll let you follow the same logic for the case of n = m. You should end up at

[math]\therefore |x| >> 0 \implies f(x) \text { approaches } \dfrac{a_0}{b_0} \ne 0 \text { because } a_0 \ne 0 \ne b_0 \text { by definition.}[/math]
To summarize this in English, if the degree of the polynomial in the numerator of a rational function is NOT greater than the degree of the polynomial in the denominator, the rational function approaches a horizontal asymptote when the absolute value of the variable becomes very large. That asymptote is the x-axis if the degree of the numerator is exceeded by the degree of the denominator. That asymptote is not the x-axis if the numerator and denominator have the same degree.

Is this clear?

Sorry for not replying sooner. I understand all of this now, thank you. y=0 when highest degree of denominator> highest degree of numerator. And for your second part at the end, you’re referring to the ratio of the coefficients of the degrees (when both degrees are equal) between the numerator and denominator. Ex: y=2/3.

Thank you all for your help!