SOLUTION: State the x-coordinate of any holes in the graph of this function: f(x)= 8x^2-16x ---------- 5x^2-20 A. 0 B. 2 C. 5 D. -2 E. None of the above

Algebra ->  Rational-functions -> SOLUTION: State the x-coordinate of any holes in the graph of this function: f(x)= 8x^2-16x ---------- 5x^2-20 A. 0 B. 2 C. 5 D. -2 E. None of the above       Log On


   

Question 1028504: State the x-coordinate of any holes in the graph of this function:
f(x)= 8x^2-16x
----------
5x^2-20
A. 0
B. 2
C. 5
D. -2
E. None of the above
Found 2 solutions by macston, robertb:
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
When the denominator is zero:
%288x%5E2-16x%29%2F%285x%5E2-20%29
5x%5E2-20=0
5x%5E2=20
x%5E2=4
x=sqrt%284%29
x=2 OR x=-2
.
I am unsure what a "hole" means, but if that is the term your instructor used,
the graph is not continuous at 2 and -2.
.
+graph%28+500%2C+500%2C+-3%2C+3%2C+-3%2C+3%2C+5x%5E2-20%29+
.
The graph should actually show a gap at 2 and -2, which is not clear in rendering.
.
Sorry, best I can do .... Mac

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
E. None of the above.
One "hole" in the graph exists at x = 2. The graph is a hyperbola with horizontal asymptote y = 8/5 and vertical asymptote x = -2.
The function can be reduced to f%28x%29+=+%288x%29%2F%285%28x%2B2%29%29, where x%3C%3E2.
The discontinuity at x = -2 cannot be considered a "hole", because it is an essential/infinite discontinuity.
Thus, there is only one "hole" in the graph.