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confusedatmath
New member
- Jan 2, 2014
- 14
The short answer, if [tex]\displaystyle \begin{align*} y = c \end{align*}[/tex] then you end up with [tex]\displaystyle \begin{align*} 0 = \frac{a}{(x + b)^2} \end{align*}[/tex]. Is that possible to solve for x?I'm stuck at understanding why +c is the horizontal asymptote. Can someone please explain this? I get that the vertical asymptote is relating to (x+b), because the denominator cannot=0. But why does this kind of graph have a horizontal asymptote to begin with?
View attachment 1838
The question was "what happens when x is large?" "2" is NOT large!the fraction gets smaller, but dont we end up adding that to the value of c??
well if a/(x-b)^2 +c
let a=1 b=1 c=2
so we get
1/(x-1)^2 +2
lets make x =2
we get
1/1 +2 .... thats 3.... wait am i doing something wrong?
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oh wait is because in the above a is negative.. so
it would be -1+2 < 2 ....