SOLUTION: 1. Showing your work at each step, analyze the graph of the following function by completing the following parts: (A)Determine the x- and y-intercepts of the graph. Write them as

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Question 468508: 1. Showing your work at each step, analyze the graph of the following function by completing the following parts:
(A)Determine the x- and y-intercepts of the graph. Write them as points.
(B)Determine the domain and vertical asymptotes, if any, of the function.
(C)Write the domain in interval notation.
(D)Determine the horizontal or oblique asymptote, if any, of the function.
(E)Obtain additional points on the graph. You should use a table to show your work in finding the additional points.
(F)Plot the x- and y-intercepts, the additional points, and the asymptotes that you found on a rectangular coordinate system and graph the function. Draw the asymptotes using dashed lines.

f(x)=x²-2x-8/x-1
Oh my goodness can someone please help me with this monster problem! PLEASE!

Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
1. Showing your work at each step, analyze the graph of the following function by completing the following parts:
(A)Determine the x- and y-intercepts of the graph. Write them as points.
(B)Determine the domain and vertical asymptotes, if any, of the function.
(C)Write the domain in interval notation.
(D)Determine the horizontal or oblique asymptote, if any, of the function.
(E)Obtain additional points on the graph. You should use a table to show your work in finding the additional points.
(F)Plot the x- and y-intercepts, the additional points, and the asymptotes that you found on a rectangular coordinate system and graph the function. Draw the asymptotes using dashed lines.
f(x)=x²-2x-8/x-1
...
I will not be able to draw you a detailed graph because I don't know how to do this with this format, but I should provide you with the information you need to draw the graph yourself.
..
f(x)=x²-2x-8/x-1
Since the numerator is one degree higher than the degree of the denominator, given function has a slant or oblique asymptote.
..
(A) Intercepts:
x-intercept: set y=0, then solve for x. In effect, you are setting the numerator=0.
x²-2x-8=0
(x-4)(x+2)=0
x-4=0
x=4
and
x+2=0
x=-2
x-intercepts: (4,0) and (-2,0)
..
y-intercept: set x=0, then solve for y
f(0)=x²-2x-8/x-1
f(0)=-8/-1=8
y-intercept: (0,8)
..
(B),(C)&(D) Domain and asymptotes:
Domain: set denominator=0, then solve for x
x-1=0
x≠1
Domain:(-∞,1) U (1,∞)
..
vertical asymptote: set denominator=0, then solve for x.
x-1=0
x=1
vertical asymptote: x=1
..
slant asymptote: by long division, divide numerator by denominator. You will get a quotient plus a remainder. The quotient is the equation of the slant asymptote, a straight line.
x²-2x-8/x-1=(x-1)+Remainder,-9/(x-1). Equation of the slant asymptote: y=x-1
There are no horizontal asymptotes
..
(E) & (F) Graphing:
See the graph below as a visual check on the answers above. It will give you a good idea what your curve should look like. You can get additional points to plot the graph by plugging in different x-values. Note that the slant asymptote is a straight line with a slope of 1 and a y-intercept of -1
..
y=(x-4)(x+2)/(x-1)

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