Question

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Answer 1

a `y=0, x=-2 and x=1`.
b Local maxima is `f(4)=1/9` and local minima is `f(0)=1`.
c x-intersept is `(2,0)` and y-intersept is `(0,1)`.
d,e See below.

Explanation:

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

a. When `x` goes to infinity, `lim_(x->oo)f(x)=0` and`lim_(x->-oo)f(x)=0` . This means that x-axis is one of the asymptotes.
In addition, `f(x)` is not defined at `x=-2,1`. Therefore `x=-2` and `x=1` are also asymptotes.

b
First, differentiate `f(x)`.
`f'(x)=((x-2)'(x^2+x-2)-(x-2)(x^2+x-2)')/(x^2+x-2)^2`
`=(1*(x^2+x-2)-(x-2)(2x+1))/(x^2+x-2)^2`
`=-(x^2-4x)/(x^2+x-2)^2`

Solving `f'(x)=0` leads to `x=0,4` and it indicates that `f(0)=1` and `f(4)=1/9` are local maxima or minima.

When `x<-2, -2< x< 0 and x>4`, `f(x)` is monotonically decreasing because `f'(x)=-(x(x-4))/(x^2+x-2)^2` is negative for this domain.
Likewise, `f(x)` is monotonically increasing when `0< x< 1 , 1< x< 4`.
Thus, `f(0)` is the local minima and `f(4)` is the local maxima.

c
Substitute `x=0` and `y=0` to the formula and the intersepts are obvious. `(2,0),(0,1)`.

d
graph{(x-2)/(x^2+x-2) [-10, 10, -5, 5]}

e
Draw the line `y=p` on the graph and find when `y=f(x)` and `y=p` cross at two distinct point. The answer is `p< 0, 0< p< 1/9 and 1< p`.