Properties of Functions - Sketching rational functions A function is defined by $f(x)=\frac{x^2+6x+12}{x+2}, x≠-2.$
a) Express f(x) in the form $ax+b+\frac{c}{x+2}$, stating the values of a, b and c.
b) Write down an equation for each of the two asymptotes.
c) Show that $f(x)$ has two stationary points. Determine the coordinates and the nature of the stationary points.
d) Sketch graph of $f$.
e) State the range of values of $k$ such that the equation $f(x)=k$ has no solution.

I have done part a and b. 
For part a) I got $x+4+\frac{4}{x+2}$. For part b) for the vertical asymptote: $x=-2$, and the horizontal asmyptote: $y=x+4$
I not't aware how you would do part $c$, $d$ and $e$. Someone please help.
 A: $a) a=1 , b=4 , c=4$ 
$b)$ Vertical asymptote: $x=−2$  and the oblique asmyptote: $ y=x+4 $ 
$c)$ Stationary points : A stationary point or critical point of a differentiable function of one variable is a point of the domain of the function where the derivative is zero (equivalently, the slope of the graph at that point is zero).
So for stationary points $\displaystyle f′(x)=\frac{x(x+4)}{(x+2)^2}=0 \Rightarrow x=0 $ or $x=-4$.
Thus $(-4,-2)$ and $(0,6)$ are coordinates of each stationary points.
At $(-4,-2)$ $f$ has a maximum and at $(0,6)$ $f$ has a minimum. Check this putting $x$ values around $-4$ and $0$
$d)$ Put $f(x)=y$. Then for $x\in \mathbb R/\{-2\}$ , $y(x+2)=x^2+6x+12 \Rightarrow x^2+(6-y)x+2(6-y)=0$
Since $x\in \mathbb R/\{-2\}$ , discriminant of this quadratic equation $\geq 0$
$\Rightarrow (6-y)^2-4\cdot 1\cdot2(6-y) \geq 0 $
$\Rightarrow (y+2)(y-6)\geq 0 $
$\Rightarrow y \in (-\infty,-2] \cup[6,+\infty) $
$\Rightarrow y$ cannot take values between  $-2$ and $6$.
Now see the graph at https://www.wolframalpha.com/input/?i=plot+%28x%5E2%2B6x%2B12%29%2F%28x%2B2%29
$e)$ 
When $f(x)=k$ , likewise in the previous case $y=k$ cannot take values between $-2$ and $ 6$ 
So if $-2<k<6 , f(x)=k$ has no solutions. 
A: hint: $$f'(x)={\frac {x \left( x+4 \right) }{ \left( x+2 \right) ^{2}}}$$
