@phi State the vertical asymptote of the rational function. \(\large\dfrac{(x-6)(x+6)}{x^2-9}\)
factor the bottom and see what you get. THere are 2 of them.
vertical asymptotes are caused by "dividing by zero" any idea what x values cause the bottom to be zero?
(x-3)(x+3)?
that is how you can see more clearly what x values make the bottom zero.
-3 and 3?
if x is -3 then (x - 3) (x+3) turns into (-3 -3) * (-3 + 3) = -6*0 = 0 and similarly for x= 3, we will get zero. when x is -3 or 3, you will get an asymptope
Ok so for vertical asymptotes I just need to factor the bottom and that will make it easy to find them?
you can remember this if you remember vertical asymptote means y "zooms up" and to get big values for y , where y = top/bottom, make the bottom *very tiny* because dividing by a tiny number gives a big number.
I have 2 more. Do you want me to put them here or open a new question?
to find the vertical asymptotes you need to find when the bottom is zero. that turns into the problem of , for example, a x^2 + b x + c = 0 i.e. solving a quadratic. for bigger degrees it gets harder!
if you post what you think the answer is.
State the domain of the rational function. \(\large\dfrac{6}{4-x}\) All real numbers except 6. All real numbers except 4. \(\Leftarrow\) I think this one All real numbers except -6 and 6. All real numbers except -4 and 4.
yes. all x's are ok except when x is 4 (which causes a divide by zero , and we don't like dividing by zero)
Give an example of a rational function that has no horizontal asymptote and a vertical asymptote at x = 1. This one I am thinking something like \(\large\dfrac{x^3+2x-8}{x^2-1}\)
from the other post, we know it does not have a horizontal one (degree of top is 3 which is different from the degree of bottom 2) and x= 1 will cause the bottom to be zero. sounds correct.
Thank you so much! I wish I could give you more medals. \(\Large\ddot\frown\)
I have lots of medals. but thanks.
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