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raytiller1
Group Title
Consider the function f(x) = x2x12/x4 . Describe the graph of this function. Include all discontinuities, intercepts, and the basic shape of the graph.
 2 years ago
 2 years ago
raytiller1 Group Title
Consider the function f(x) = x2x12/x4 . Describe the graph of this function. Include all discontinuities, intercepts, and the basic shape of the graph.
 2 years ago
 2 years ago

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mathmate Group TitleBest ResponseYou've already chosen the best response.1
Consider the expression: (x^2x12)/(x4)=(x4)(x+3)/(x4) =(x3) after cancelling the common factor for points where x not equal to 3. At x=3, the function is undefined. Is that enough to get you gonig?
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
You can't cancel and forget about the x4 term. There's still a point discontinuity there.
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
But otherwise it behaves like the line.
 2 years ago

raytiller1 Group TitleBest ResponseYou've already chosen the best response.0
now im really confused @mathmate @vf321
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
Sorry, I meant to say that the function is undefined at x=4 (where we cancelled). The remaining points behave like y=x+3.
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
@raytiller1 What mathmate's saying is right  If you notice, \[f(x)=\frac{x^2x12}{x4} = \frac{(x4)(x+3)}{x4}\] So, when x!=4, you have some number x4 on the numerator and in the denominator, so they cancel. \[x \neq 4::f(x) = \frac{x4}{x4}(x+3)=1(x+3)\] When x does equal 4, you get 0/0, a discontinuity.
 2 years ago

raytiller1 Group TitleBest ResponseYou've already chosen the best response.0
thanks guys @vf321 @mathmate
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
You're welcome! :)
 2 years ago
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