This is for review for my final exam. I would like to learn how to grasp the concept and answer the problem correctly, so please explain each step.
Equation: f(x)=5x^2-x over x-2
Find the equation of the horizontal and vertical asymptotes, if any.
Find the equation of slant asymptotes (if any) ..
Thank you in advance & will choose best answer!
Equation: f(x)=5x^2-x over x-2
Find the equation of the horizontal and vertical asymptotes, if any.
Find the equation of slant asymptotes (if any) ..
Thank you in advance & will choose best answer!
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For vertical asymptotes, look for places where the denominator is 0
In this case you have vertical asymptotes at x=2.
To sketch a graph of it you'll want to find the sign of the function near the pole, at x=2.01 and x=2.99 for example, or use inequalities to determine the sign of 5x^2-x there. (it's positive. if x->2+ from the right that is then you get +oo and from x->2- it's -oo.)
For horizontal asymptotes, that happens when the limit of the function as x->oo is a constant or x->-oo
When I type oo that means the infinity symbol which is otherwise troublesome to type, sorry.
For slant asymptotes, it seems to me that one way is to find the limit of f(x)/x as x->+oo or x->-oo and that would be the slope of it, if you get a constant for the limit. Then trying to find the intercept of it seems somewhat impossible, but it must be possible, so I have to look it up or think harder about it ....
well the thing to do is find a line g(x)=ax+b so lim (x->oo) f(x) - g(x) = 0.
I think the process can be to find a first and then find lim(f(x) - ax) = b.
a is equal to the limit of f(x)/x since b is a real number.
For this one
lim (x->oo) = +oo
lim(x->-oo) = -oo
lim(x->oo) (5x^2 - x) / (x(x-2)) = 5
same at -oo
lim(x->oo) (5x^2 - x)/(5x(x-2)) = (5x^2 - x)/(5x^2 - 10x)
= 1 + 9x/(5x^2 - 10x) because -1 = -10+9
this has limit +1 as x->+oo or -oo
so the slant asymptote is y=5x+1.
In this case you have vertical asymptotes at x=2.
To sketch a graph of it you'll want to find the sign of the function near the pole, at x=2.01 and x=2.99 for example, or use inequalities to determine the sign of 5x^2-x there. (it's positive. if x->2+ from the right that is then you get +oo and from x->2- it's -oo.)
For horizontal asymptotes, that happens when the limit of the function as x->oo is a constant or x->-oo
When I type oo that means the infinity symbol which is otherwise troublesome to type, sorry.
For slant asymptotes, it seems to me that one way is to find the limit of f(x)/x as x->+oo or x->-oo and that would be the slope of it, if you get a constant for the limit. Then trying to find the intercept of it seems somewhat impossible, but it must be possible, so I have to look it up or think harder about it ....
well the thing to do is find a line g(x)=ax+b so lim (x->oo) f(x) - g(x) = 0.
I think the process can be to find a first and then find lim(f(x) - ax) = b.
a is equal to the limit of f(x)/x since b is a real number.
For this one
lim (x->oo) = +oo
lim(x->-oo) = -oo
lim(x->oo) (5x^2 - x) / (x(x-2)) = 5
same at -oo
lim(x->oo) (5x^2 - x)/(5x(x-2)) = (5x^2 - x)/(5x^2 - 10x)
= 1 + 9x/(5x^2 - 10x) because -1 = -10+9
this has limit +1 as x->+oo or -oo
so the slant asymptote is y=5x+1.