Rational Functions...1

[nycmathguy]

Section 2.6
Question 26

In Exercises 17–38, (a) state
the domain of the function, (b) identify all
intercepts, (c) find any vertical or horizontal
asymptotes, and (d) plot additional solution
points as needed to sketch the graph of the rational function.

Note: Part (d) will NOT be done by hand.

Question 26

Part (a)

Domain: All real numbers except for x = 2.

Part (b)

Note: f(x) = y.

Let x = 0

f(0) = -0/(0 - 2)^2

f(0) = -0/4 = 0

f(0) = 0 = y

The y-intercept is the line y = 0 aka the x-axis.

Let f(x) = 0

0 = -x/(x - 2)^2

0 = -x

0/-1 = x

0 = x

The x-intercept is the line x = 0 aka the y-axis.

Part (c)

Top degree < bottom degree.
The horizontal asymptote is the line y = 0.

NOTE: Notice that the line y = 0 is also the x-axis as stated in Part (b).

Part (d)

 

[MathLover1]

part (c)
Top degree < bottom degree.
The horizontal asymptote is the line y = 0 =>correct
vertical asymptote is x=2

(d) plot additional solution points means you make table, find f(x) for chosen values of x, then plot them and draw a graph

 

[nycmathguy]

part (c)
Top degree < bottom degree.
The horizontal asymptote is the line y = 0 =>correct
vertical asymptote is x=2

(d) plot additional solution points means you make table, find f(x) for chosen values of x, then plot them and draw a graph

1. Can we say the domain is the same as the vertical asymptote at x = 2?

2. I understand what Part (d) is asking but my time is limited in terms of graphing by hand. I will be using Desmos for most graph work.

 

[MathLover1]

1. yes, the domain is the same as the vertical asymptote at x = 2

 

[nycmathguy]

1. yes, the domain is the same as the vertical asymptote at x = 2

Very good. We continue with Section 2.6 later today, tomorrow and Monday morning.