## Show full working out please

Oct 20, 2017

a $y = 0 , x = - 2 \mathmr{and} x = 1$.
b Local maxima is $f \left(4\right) = \frac{1}{9}$ and local minima is $f \left(0\right) = 1$.
c x-intersept is $\left(2 , 0\right)$ and y-intersept is $\left(0 , 1\right)$.
d,e See below.

#### Explanation:

$f \left(x\right) = \frac{x - 2}{{x}^{2} + x - 2} = \frac{x - 2}{\left(x + 2\right) \left(x - 1\right)}$

a. When $x$ goes to infinity, ${\lim}_{x \to \infty} f \left(x\right) = 0$ and${\lim}_{x \to - \infty} f \left(x\right) = 0$ . This means that x-axis is one of the asymptotes.
In addition, $f \left(x\right)$ is not defined at $x = - 2 , 1$. Therefore $x = - 2$ and $x = 1$ are also asymptotes.

b
First, differentiate $f \left(x\right)$.
$f ' \left(x\right) = \frac{\left(x - 2\right) ' \left({x}^{2} + x - 2\right) - \left(x - 2\right) \left({x}^{2} + x - 2\right) '}{{x}^{2} + x - 2} ^ 2$
$= \frac{1 \cdot \left({x}^{2} + x - 2\right) - \left(x - 2\right) \left(2 x + 1\right)}{{x}^{2} + x - 2} ^ 2$
$= - \frac{{x}^{2} - 4 x}{{x}^{2} + x - 2} ^ 2$

Solving $f ' \left(x\right) = 0$ leads to $x = 0 , 4$ and it indicates that $f \left(0\right) = 1$ and $f \left(4\right) = \frac{1}{9}$ are local maxima or minima.

When $x < - 2 , - 2 < x < 0 \mathmr{and} x > 4$, $f \left(x\right)$ is monotonically decreasing because $f ' \left(x\right) = - \frac{x \left(x - 4\right)}{{x}^{2} + x - 2} ^ 2$ is negative for this domain.
Likewise, $f \left(x\right)$ is monotonically increasing when $0 < x < 1 , 1 < x < 4$.
Thus, $f \left(0\right)$ is the local minima and $f \left(4\right)$ is the local maxima.

c
Substitute $x = 0$ and $y = 0$ to the formula and the intersepts are obvious. $\left(2 , 0\right) , \left(0 , 1\right)$.

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

e
Draw the line $y = p$ on the graph and find when $y = f \left(x\right)$ and $y = p$ cross at two distinct point. The answer is $p < 0 , 0 < p < \frac{1}{9} \mathmr{and} 1 < p$.