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primeralph
 one year ago
Best ResponseYou've already chosen the best response.0talki to @DrPepperx3 I thin she's doing the same problems

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1These problems just have vertical asymptotes. To find a vertical asymptote take the function at the bottom and equal it to 0. When the denominator of a function equals 0, or does not exist, you have a vertical asymptote.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1here is a site that helped me understand asymptotes http://www.purplemath.com/modules/asymtote4.htm

ValentinaT
 one year ago
Best ResponseYou've already chosen the best response.0Okay, thanks! When I'm finished finding the asymptotes, could you help me with graphing them? If you don't mind.

ValentinaT
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so I got: 36. x = 2, y = 0 37. x = 3, y = 0 38. x = 4, y = 1 39. x = 5, y = 1

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1So when you're graphing vertical asymptotes for problems 3639, all you need to do is use your xintercepts and draw a line that crosses them. a "vertical" asymptote will intersect only one xvalue and it means that the function will come really close to that point but it will never cross it.

ValentinaT
 one year ago
Best ResponseYou've already chosen the best response.0Okay, I finished graphing 36  39. Could you help me with 40?

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1welfor part b, i believe the intensity of the light wouldnt be as strong, considering you're further away from the light bulb.
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