Posted by **angie** on Tuesday, January 10, 2012 at 1:04pm.

find the domain and range of the function f(x)=x^2-9/4x^2+x

- math/calculus -
**Steve**, Tuesday, January 10, 2012 at 3:28pm
You have

(x-3)(x+3) / x(4x+1)

The domain is all real numbers except where the denominator is zero. So, that would be all reals except 0 and -1/4

There is a horizontal asymptote at y = 1/4.

There are vertical asymptotes at x = -1/4 and x=0.

For x between -1/4 and 0, y > 0

As x gets big, y lies below the asymptote.

So, the range outside (-1/4,0) is all reals less than 1/4.

y' = (x^2+72x+9)/(4x^2+1)^2

y' = 0 when x = -.125

y(-.125) = 143.75

So, the range is all reals

< 1/4 or >= 143.75

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