vertical asymptotes is easy because
x = 3 or -3 makes function undefined
where is slope + and where - ?
slope = ([x^2-9]2x - x^2[2x])/[x^2-9]^2
denominator is always + or 0
when is top + and -?
top is -18x
so slope = -18x/positive number
so
slope is + if x is + and - if x is -
do next derivative same way, when numerator is + that is a bottom and when - that is a top
Consider the following function f(x)=x^2/[x^2-9]
f(x) is increasing on the interval(s)
f(x) is decreasing on the interval(s)
f(x) has 2 vertical asymptotes x=
f(x) is concave up on the interval(s)
f(x) concave down on the interval(s)
I've been stuck on these parts, I answered every other question like this, but cannot figure out this one.
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