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

Answer 1

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