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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.
 one year ago
 one year 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.
 one year ago
 one year 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?
 one year 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.
 one year ago

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

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

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

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