Question

For f(X)= 3x^3-3x^2-6x / 2x^5-2x ,
which of the following is NOT true of f?

Answer 1

x = 1 is an asymptote of f

x = -1 is NOT an asymptote of f

f has a root at x = 2

the axis is an asymptote of f

the y-axis is an asymptote of f

Answer 2

I think option d means 'x-axis' but it wasn't there in the problem.

I know that a, b, and c are true
but I can't narrow down the rest

Answer 3

for the graph of this equation, I know that as x-> +/- infinity, f(x)->0
and as x->1 f(x)-> +/- infinity
but then which axis is the asymptote?

Answer 4

i had d as my answer but changed it to e but i'm still not sure. please help!

Answer 5

You MUST remember your Order of Operations.

f(x)= 3x^3-3x^2-6x / 2x^5-2x = \(3x^{3} - 3x^{2} - \dfrac{6x}{2x^{5}} - 2x\). Off the top of my head, I'm guessing this is NOT what you intended. It is so simple to add proper grouping symbols.

f(x)= (3x^3-3x^2-6x) / (2x^5-2x) = \(\dfrac{3x^{3} - 3x^{2} - 6x}{2x^{5} - 2x}\)

Now you can simplify: \(f(x) = \dfrac{3(x^{3} - x^{2} - 2x)}{2(x^{5} - x)} = \dfrac{3x(x^{2} - x - 2)}{2x(x^{4} - 1)}\)

And factor more: \(f(x) = \dfrac{3x(x-2)(x+1)}{2x(x^{2} - 1)(x^{2}+1)} = \dfrac{3x(x-2)(x+1)}{2x(x+1)(x-1)(x^{2}+1)}\)

Now, we can start answering questions. HOLES!! \(x\ne 0\;and\;x\ne -1\). With that understanding, we can simplify some more.

\(f(x) = \dfrac{3(x-2)}{2(x-1)(x^{2}+1)}\)

Vertical Asymptote: x = 1
Horizontal Asymptote: y = 0

Don't try to guess or work piece-meal. Tackle it and do it up right!

Answer 6

Thanks for such a detailed reply! So because the horizontal asymptote is y=0, that means that d is true, and e is the answer because it is NOT true?

Answer 7

D doesn't mean anything. Should that be the x-axis?

YES x = 1 is an asymptote of f

YES x = -1 is NOT an asymptote of f

YES f has a root at x = 2, f(2) = 0

UNCLEAR the axis is an asymptote of f
YES the x-axis is an asymptote of f

NO the y-axis is an asymptote of f

Answer 8

i'm pretty sure D just had a spelling error, yes. So i'll put e as my answer! Thanks very much for your help!