# Homework Help: Rational Expressions loooong BUT NEED HELP

1. Dec 8, 2004

### aisha

state the vertical asymptote(s), x-intercept(s), y-intercept(s), domain and range of f(x)=(1)/(4x^2)-1

ok I factored the denominator and got (2x+1) (2x-1) I solved for x, so the x-intercepts are x=-1/2, and x=1/2 for only (4x^2)-1 the reciprocal function has no x-intercepts.

Sub in x=0 to get y intercept, I got y=-1 for (4x^2)-1
that means for f(x)=(1)/(4x^2)-1 it must be the same since -1 is an invariant point.

Ok I think the vertex is (0,-1) for (4x^2)-1 this is also the vertex of f(x)=(1)/(4x^2)-1 but this function has a maximum instead of minimum.

I gather the vertical asymptotes are x=-0.5, and x=0.5 for f(x)=(1)/(4x^2)-1

The Range for f(x)=(1)/(4x^2)-1 {y:y does not equal 0 but is YER} if this is correct what is the proper notation.

The Domain for f(x)=(1)/(4x^2)-1 {x:does not equal -1/2 or 1/2, but is XER}
if this is correct what is the proper notation?

Ok um I dont know if all that I have said is correct, can someone please Help me out?

I have a sketch of the graph but the forum wont let me attach it it is saying it is too big....

it sort of looks like JUL
n
a parabola with a min is the original and then the reciprocal is a parabola with a max and in the top two quadrants a backwards L and L .
I DONT KNOW IF THIS MAKES ANY SENSE, but I TRIED.

Last edited: Dec 8, 2004
2. Dec 9, 2004

### Galileo

Hi Aisha,

Is the function
$$f(x)=\frac{1}{4x^2}-1$$ or $$f(x)=\frac{1}{4x^2-1}$$?

Your notation implied the former one, but then you cannot factor the denominator like $(2x-1)(2x+1)$. You can do so only in the latter one. I take it that's the one you meant.

The notation you want for the range is :$$\mathbb{R}\setminus \{0\}$$, which means $\mathbb{R}$ minus the set $\{0\}$. So minus the point 0.
It's not correct though.
The function looks more like $J_\cap L[/tex] then JUL :rofl: Anyway, look for the extrema, you`ll find there's a local maximum at x=0. (0,-1), I guess you already found it. Never heard it being called a vertex though. That means the function is smaller than -1 in the interval (-1/2,1/2) and larger than 0 elsewhere in its domain. The domain is correct, a correct notation would be [itex]\mathbb{R}\setminus \{-\frac{1}{2},\frac{1}{2}\}$

The asymptotes are ok too.

3. Dec 11, 2004

### aisha

Yes It Is The Second One!! I Still Need More Help Please!!! Anyone!!