pre calc help important!!
state the vertical asymptote of the rational function. f(x) = (x-3)(x+4)/x^2-1
\(\large\color{black}{ f(x)= \frac{\LARGE (x-3)(x+4)}{\LARGE x^2-1} }\)
right?
yes
So, we are just doing a basic observation of the domain. You know that: \(\large\color{black}{ x^2-1 \ne 0 }\) right?
yes
so what is the restriction for x. What x cannot equal?
x can't equal 0?
\(\large\color{black}{ x^2-1=0 }\) solve for x.
x^2=1
and x is equal to?
x=3 and x=-4
wait,. if x^2=1, then x=? (there are 2 solutions)
these solutions are the vertical asymptotes.
take the square root fo each side.
what do you mean square root of each side why wouldn't the answer be 3 and -4?
\(\large\color{black}{ x^2-1 =0 }\) \(\large\color{black}{ x^2=1 }\) \(\large\color{black}{ x=? }\)
(we are not finding when the function is a zero, we are finding the undefined values for the denominator)
oh +/-1
Yes. \(\large\color{black}{ \pm 1 }\)
thanks! can u help me with 1 more problem
So your vertical asymptotes are: \(\large\color{black}{ x=1 }\) and \(\large\color{black}{ x=-1 }\)
I can try;)
but in a different question.
why a different one? why not just on here
Well, I don't mind, but the rules of this website are that you can ask as many questions as you need, but you should ask one question per post.
One reason I can think of, is thta the thread shouldn't have a very slow loading time.
you can ask it here for now though
State the horizontal asymptote of the rational function. f(x) = x+9/x^2+2x+3
I wouldn't call x=0 an asymptote, well, take a look here. https://www.desmos.com/calculator/iukundkwss
i'm solving for y
I think it is just y=0, if any horizontal asymptote.
I meant y initially, but wrote x-:(
oh ok. the answer choices are y = x None y = 0 y = 9
y=0 seems to be it
ok thanks
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