if range of $f(x) = \frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4]$. Then $a$ and $b$ are If Range of $\displaystyle f(x) = \frac{x^2+ax+b}{x^2+2x+3}$ is $\left[-5,4\; \right]$ for all $\bf{x\in \mathbb{R}}$. Then values of $a$ and $b$.
$\bf{My\; Try}::$ Let $\displaystyle y=f(x) = \frac{x^2+ax+b}{x^2+2x+3} = k$,where $k\in \mathbb{R}$.Then $\displaystyle kx^2+2kx+3k=x^2+ax+b$
$\Rightarrow (k-1)x^2+(2k-a)x+(3k-b) = 0$
Now we will form $2$ cases::
$\bf{\bullet}$ If $(k-1)=0\Rightarrow k=1$, Then equation is $(2-a)x+(3-b)=0$
$\bf{\bullet}$ If $(k-1)\neq 0\Rightarrow k\neq 1$ means either $k>1$ or $k<1$
How can i solve after that
Help Required
Thanks 
 A: Note that $x^2+2x+3\gt 0 $ for any $x\in\mathbb R$.
Hence, we have
$$-5(x^2+2x+3)\le x^2+ax+b\le 4(x^2+2x+3).$$
You can simplify this, and you have two quadratic inequality which will be easy to solve.
You'll have two inequality.
$$6x^2+(a+10)x+b+15\ge0$$
$$3x^2+(8-a)x+12-b\ge0.$$
This leads that each discriminant has to be equal or greater than $0$. So, we have
$$(a+10)^2-4\cdot 6\cdot (b+15)\le 0$$
$$(8-a)^2-4\cdot 3\cdot (12-b)\le 0.$$ 
A: To find the places where $f(x)$ is minimal and maximal, differentiate $f$ wrt $x$. Then solve $f'(x)=0$. Call the solution $x_0$ an $x_1$ (and so on if there are move). Now, you know for which $x$ $f(x)$ is minimal/maximal. Calculate $f(x_0)$ and $f(x_1)$. These should be equal to $-5$ and $4$. You only have to know which one is a minimum and which one a maximum, but you should be able to figure that out.
A: We need $$(2k-a)^2-4(k-1)(3k-b)=-8k^2+(b+3-4a)k+a^2-4b\ge0$$
$$\iff 8k^2-(b+3-4a)k-(a^2-4b)\le0$$
Now, we know $\displaystyle (x-a)(x-b)\le0\iff a\le x\le b$ where $a\le b$
So, here  $$x^2-(-5+4)x+(-5)4=0\iff x^2+x-20=0\  \ \    \  (1)$$ will be same as $$8k^2-(b+3-4a)k-(a^2-4b)=0\  \ \ \ (2)$$
Now use Find $k, m$ so solutions to $(iz+k)^2=-2+2\sqrt3i$ are the same as those to $z^2-2iz+m=0$ to find $a,b$
