Rational Function With Slant Asymptote...2

Section 2.6
Question 60




Before doing anything, I simplified question 60 to become f(x) = (2x^2 + 5x + 2)/(x - 1).

Part (a)

Domain = All real numbers except for x = 1.

Part (b)

The x-intercepts are x = -2, x = -1/2 crossing through the points (-2, 0) and (-1/2, 0).

The y-intercept is the line y = -2 and located at the point (0, -2).

Part (c)

The vertical asymptote is the line x = 1.
The slant asymptote is the binomial 2x + 7.

Part (d)

I created the following table:

x..............f(x)

-1.............1/2
-2.............0
0..............-2
2..............20
3..............35/2

Before drawing the graph of the function, is the table of points correct? Is the slant asymptote correct? Is the vertical asymptote correct? Are the intercepts correct? Do I graph the slant asymptote by graphing the dotted line y = 2x + 7?

 

you simplified question 60 to become f(x) = (2x^2 + 5x + 2)/(x - 1)

Part (a)
Domain = All real numbers except for x = 1.-> correct

but when you simplified you lost (x-2)

denominator is x^2-3x+2= (x - 1) (x - 2)=> Domain = All real numbers except for x = 1 and x=2

Part (b) correct

Part (c)




Part (d)

I created the following table:

x..............f(x)

-1.............1/2 -> correct
-2.............0 -> correct
0..............-2-> correct
2..............20 -> incorrect ( x=2 is excluded in domain because f(x) is undefined)
3..............35/2 -> correct

 

Look left part (c) blank. What is part (c)?

 

your answer in part (c) is correct

 

From what I can see, I made a few minor errors.