Rational Function With Slant Asymptote...1

Section 2.6
Question 59

Can you do 59 (a-d) as a guide for me to do a few more on my own? Please, show the steps.

 

f(x)=(2x^3-x^2-2x+1)/(x^2+3x+2)

a.
domain
exclude values of x that make x^2+3x+2=0 which are x=-2 and x=-1

then domain is all real numbers
x<-2 or -2<x<-1 or x>-1
interval notation:
(-infinity , -2 ) U (-2, -1 ) U (-1, infinity )

b.
set f(x)=0 to find x intercepts
0=(2x^3-x^2-2x+1)/(x^2+3x+2)
((x - 1) (2 x - 1))/(x + 2) = 0

will be 0 only if 0=(x - 1) (2 x - 1)
and it is
=> x=-1, x=1/2
x intercepts: ( 1/2,0), (1,0)

set x=0 to find y intercepts
y=(2*0^3-0^2-2*0+1)/(0^2+3*0+2) =1/2

y intercept0,1/2)


c. asymptotes

vertical: x=-2,
horizontal:
A rational function has a slant asymptote if the degree of a numerator polynomial is 1 more than the degree of the denominator polynomial.

f(x)=(2x^3-x^2-2x+1)/(x^2+3x+2) use long division
2x^3 - x^2 - 2 x + 1 = ((2 x - 7))*(x^2 + 3 x + 2) + (15 x + 15)
so, quotient is 2 x - 7
and horizontal asymptote is: y=2x-7 (slant )

d.
plot additional points

x | f(x)
-3| -7 ..........point (-3,-7)
-1| 0..........point (-1,0)
0| 1/2 ........point (0,1/2)
1| 0 .........point (1,0)
2| 9/8 .......point (2,9/8)
you can find more points if you want , then draw a graph with slant asymptote y=2x-7

 

Why did you post a sad face next to the y-intercept (0,1/2)?

You said:

"horizontal asymptote is: y=2x-7 (slant )"

Of course you meant slant asymptote, right?
I can plot the slant asymptote 2x - 7 by making a table of points. Yes?

 

probably came because I did not make a space between y intercept and (0,1/2)

yes, I meant slant asymptote

yes, making a table of two points is enough to plot them and draw a line through

 

Very good. Just moving along the precalculus way trying to make sense of the current section. There's a lot to intake. Why is that? Because I am trying to do too much at the same time. I promised not to post so many daily questions but find myself often breaking my own promises.