SOLUTION: really need this answer 1)Graph f(x)=(x^2 - 4)/(x - 2) using a graphing calculator. Using a standard window with the trace feature, trace the graph to x=2. What happens? 2.

Algebra ->  Rational-functions -> SOLUTION: really need this answer 1)Graph f(x)=(x^2 - 4)/(x - 2) using a graphing calculator. Using a standard window with the trace feature, trace the graph to x=2. What happens? 2.      Log On


   

Question 79627: really need this answer
1)Graph f(x)=(x^2 - 4)/(x - 2) using a graphing calculator. Using a standard window with the trace feature, trace the graph to x=2. What happens?
2. Why is there a "break" in the line at x=2? Why is the graph a line without a vertical asymptote at x=2? If x is "close to" 2, what is the y value "close to"?
Create a function that has either a hole, a break, or an asymptote in the graph.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1)Graph f(x)=(x^2 - 4)/(x - 2) using a graphing calculator. Using a standard window with the trace feature, trace the graph to x=2. What happens?
graph%28400%2C300%2C-10%2C10%2C-10%2C10%2C%28x%5E2-4%29%2F%28x-2%29%29
At x=2 you get an ERROR.
2. Why is there a "break" in the line at x=2?
The denominator cannot be zero; Because there is also a factor of
(x-2) in the numerator you get a "hole" in the graph.
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Why is the graph a line without a vertical asymptote at x=2?
If you cancel the (x-2) factor that is common to numerator and denominator
you get the function y=x+2 which graphs as a line.
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If x is "close to" 2, what is the y value "close to"?
y=4
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Create a function that has either a hole, a break, or an asymptote in the graph.
Asymptote: y=x/(x-2) will have a vertical asymptote at x=2.
Hole: y=[x^2-3x]/[x-3] will have a hole at x=3
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Cheers,
Stan H.