state the vertical asymptote(s), x-intercept(s), y-intercept(s), domain and range of f(x)=(1)/(4x^2)-1(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Rational Expressions loooong BUT NEED HELP

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