Rational Function With Slant Asymptote...3

[nycmathguy]

Section 2.6
Question 52

Part (a)

x - 1 = 0

x = 1

Domain = All real numbers except for x = 1.

Part (b)

h(x) = x^2/(x - 1)

Let h(x) = 0

0 = x^2/(x - 1)

0(x - 1) = x^2

0 = x^2

0 = x

h(0) = 0^2/(0 - 1)

h(0) = 0/-1

h(0) = 0

The x-intercept = 0 = y-intercept.

Part (c)

The vertical asymptote is x = 1.
The slant asymptote is the line h(x) = x + 1.

Part (d)

x............h(x)
0.............1
-1............0

Before graphing the given function, is my work here right?

 

[MathLover1]

a, b, and c are correct

Part (d)
should be
x............h(x)
0.............0 ->h(0) = 0^2/(0 - 1)=0/-1=0
-1............-1/2 ->h(-1) = (-1)^2/(-1 - 1) =1/-2=-1/2

 

[nycmathguy]

a, b, and c are correct

Part (d)
should be
x............h(x)
0.............0 ->h(0) = 0^2/(0 - 1)=0/-1=0
-1............-1/2 ->h(-1) = (-1)^2/(-1 - 1) =1/-2=-1/2

Thank you. I will post the graph tomorrow. On the 4 train now to work. Maybe some word problems later. We are done with Section 2.6.

 

[nycmathguy]

a, b, and c are correct

Part (d)
should be
x............h(x)
0.............0 ->h(0) = 0^2/(0 - 1)=0/-1=0
-1............-1/2 ->h(-1) = (-1)^2/(-1 - 1) =1/-2=-1/2

Here are some more additional points to make our graph:

x.........h(x)
-2.........-4/3
-1.........-1/2
0..............0
2..............4

 

[MathLover1]

correct

 

[nycmathguy]

correct

Wish I had more time to graph by hand. Lots of fun for sure.