SOLUTION: Which function best describes a function that had a point of discontinuity of 1/2 and a vertical asymptote of x=3 A) 2x^2-x/2x^2-7x+3 B) 2x^2-7x+3/2x^2-x C) 2x^2-1/2x^2+7x+3 D

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Which function best describes a function that had a point of discontinuity of 1/2 and a vertical asymptote of x=3 A) 2x^2-x/2x^2-7x+3 B) 2x^2-7x+3/2x^2-x C) 2x^2-1/2x^2+7x+3 D      Log On


   

Question 1047840: Which function best describes a function that had a point of discontinuity of 1/2 and a vertical asymptote of x=3
A) 2x^2-x/2x^2-7x+3
B) 2x^2-7x+3/2x^2-x
C) 2x^2-1/2x^2+7x+3
D) 2x^2+7x+3/2x-1
Found 2 solutions by Boreal, ewatrrr:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Vertical asymptote requires denominator be 0.
2x^2-7x+3=0 for x=3
2x^2-x equals 15
2x^2+7x+3=42
and 2x-1=7
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C%282x%5E2-x%29%2F%282x%5E2-7x%2B3%29%29
to show the discontinuity, factor this
(x(2x-1))/(2x-1)(x-3)
the (2x-1) will factor out, but graphing, the denominator will be 0 when x=(1/2). This is different from the -3 asymptote, because here, as the function approaches 1/2 from both sides, it has the same slope. There is a hole in the function only at 1/2.
The answer is A.

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
A) 2x^2-x/(2x^2-7x+3)
A) 2x^2-1/(2x - 1)(x-3)
Denominator CANNOT be ZERO
x = 1/2 (pt of discontinuity) and x= 3 (asymptote)