You need to notice that the function will have two vertical asymptotes at `x = +-1` , since the function is not defined at `x = +-1.`

You need to check if the function has maximum and minimum points, hence, you need to evaluate the derivative of the function, such that:

`f'(x) = ((2x^2 - 5x)'(x^2 - 1) - (2x^2 - 5x)(x^2 - 1)')/((x^2 - 1)^2)`

`f'(x) = ((4x - 5)(x^2 - 1) - 2x(2x^2 - 5x))/((x^2 - 1)^2)`

`f'(x) = (4x^3 - 4x - 5x^2 + 5 - 4x^3 + 10x^2)/((x^2 - 1)^2)`

Reducing duplicate terms yields:

`f'(x) = (5x^2 - 4x + 5)/((x^2 - 1)^2)`

You should notice that `f'(x) = 0` if` 5x^2 - 4x + 5 = 0` and since `5x^2 - 4x + 5 > 0` for all `x in R - {+-1}` , the function will not have maximum or minimum points.

You also may notice that `f'(x) > 0` for `x in` `R - {+-1}` , hence, the function will strictly increase over `R - {+-1}.`

You need to check if the function has inflection points, hence, you need to solve for x the equation `f''(x) = 0` , but fisrt you need to evaluate `f''(x)` , such that:

`f''(x) = ((5x^2 - 4x + 5)'((x^2 - 1)^2) - (5x^2 - 4x + 5)((x^2 - 1)^2)')/((x^2 - 1)^4)`

`f''(x) = ((10x - 4)((x^2 - 1)^2) - 4x(x^2 - 1)(5x^2 - 4x + 5))/((x^2 - 1)^4)`

Factoring out `(x^2 - 1)` yields:

`f''(x) = ((x^2 - 1)(10x - 4)((x^2 - 1) - 4x(5x^2 - 4x + 5)))/((x^2 - 1)^4)`

Reducing duplicate factors yields:

`f''(x) = ((10x - 4)(x^2 - 1) - 4x(5x^2 - 4x + 5))/((x^2 - 1)^3)`

`f''(x) = (10x^3 - 10x - 4x^2 + 4 - 20x^3 + 16x^2 - 20x)/((x^2 - 1)^3)`

`f''(x) = (-10x^3 + 12x^2 - 30x + 4)/((x^2 - 1)^3)`

The equation `f''(x) = 0` has one real root at `x in (0,1)` since `f''(0) = 4 > 0` and `f''(1) = -24 < 0.`

You also should notice that the second order derivative increases over `(-oo,-1)U(0.14,1)` and decreases over `(-1,0.14)U(1,oo).`

Sketching the graph of the given function yields:

Graph `f(x)=(2x^2-5x)/(x^2-1)` :

Without calculus:

(1) Vertical asymptotes -- a rational function has vertical asymptotes if the function is written without common factors in the numerator and denominator, and the denominator is zero for some value(s) of x.

Here there are no common factors in the numerator and denominator and the denominator is 0 at x=1 and x=-1. There are vertical asymptotes at x=-1 and x=1.

(2) Horizontal asymptotes -- since the degree of the numerator and denominator are the same there will be a horizontal asymptote at `y=a_n/b_n` where `a_n,b_n` are the leading coefficients of the numerator and denominator when written in standard form.

Here `a_n=2,b_n=1` so there is a horizontal asymptote at y=2.

(3) The y-intercept occurs when x=0 and is y=0.

(4) Any x-intercepts occur when the numerator is zero and the denominator is nonzero at those values for x.

`2x^2-5x=0 ==> x(2x-5)=0 ==> x=0` or `x=5/2` .

There are two x-intercepts; `x=0,x=5/2`

(5) Now try some points in the intervals x<-1, -1<x<0, 0<x<1,`1<x<5/2,5/2<x` .

On x<-1 the function is always positive.

On -1<x<0 the function is negative.

On 0<x<1 the function is positive.

On `1<x<5/2` the function is negative.

On `5/2<x` the function is positive.

(Try test values like x=-3,x=-2,x=-1/2,x=1/2,x=2,x=4 etc...)

The graph:

(The red dotted line is not part of the graph -- it is there to indicate the horizontal asymptote.)