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anonymous
 one year ago
How do I identify the intercepts, removable continuities, and asymptotes?
(2x^33x+1)/x^35x+7
anonymous
 one year ago
How do I identify the intercepts, removable continuities, and asymptotes? (2x^33x+1)/x^35x+7

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1are you allowed to graph the function ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nope :/ i think i'm supposed to solve it based on just the equation

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Finding horizontal asymptotes is easy for a rational function

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[y = \dfrac{\color{Red}{2}x^33x+1}{\color{red}{1}x^35x+7}\] As \(x\) gets large, the lower degree terms don't matter much, so the value approaches \[y \approx \dfrac{\color{Red}{2}x^3}{\color{red}{1}x^3} = \dfrac{\color{Red}{2}}{\color{red}{1}}=\color{red}{2}\] so \(\color{red}{y=2}\) is an horizontal asymptote

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you!! is the vertical asymptote about 2.75?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x^35x=7 right? and solve for x

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1it should be around x= 2.75 but how did u get that ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay thanks!! i'm super confused on finding the x intercepts though

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1did you find y intercepts ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i got 1/7 by plugging in 0 in place of x

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Good. use the same trick simply plugin y=0 to get the x intercepts

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[y = \dfrac{\color{black}{2}x^33x+1}{\color{black}{1}x^35x+7}\] plugin y=0 for x intercepts \[0 = \dfrac{\color{black}{2}x^33x+1}{\color{black}{1}x^35x+7}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1which is same as \[0 =\color{black}{2}x^33x+1\] solve

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1wolfram gives 3 x intercepts http://www.wolframalpha.com/input/?i=solve+2x%5E33x%2B1%3D0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0could i use the formula bc of the plus and minus?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1nope, the equation is cubic, not quadratic. i would simply use calculator/wolfram

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm confused on how i would get 3 answers :/

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1why ? it is fine to have multiple x intercepts

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that just means, the graph cuts the x axis at 3 different places

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i used mathway and it gave me (1,0) and (1+sqrt3)/2 but i'm not sure how it got that

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437717579760:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh i see it haha thank you so much :)
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