We claim that it is impossible for the function $f(x)=x^2+5x+4$ to be sent to $g(x)=x^2+10x+8$ using any combination of the given transformations. Without loss of generality, we will only consider the transformations of the quadratic function $ax^2+bx+c$. If the transformation of the first kind is applied to $ax^2+bx+c$, we obtain $x^2\left(a\left(\frac{1}{x}+1\right)^2+b\left(\frac{1}{x}+1\right)+c\right)=(a+b+c)x^2+(2a+b)x+a$. If the transformation of the second kind is applied to $ax^2+bx+c$, we obtain $(x-1)^2\left(\frac{a}{(x-1)^2}+\frac{b}{x-1}+c\right)=cx^2+(b-2c)x+(a-b+c)$. Now, observe that both of these transformations preserve the determinant of the original function. However, as the determinant of $f(x)$ is not equal to that of $g(x)$, it is impossible to transform $f(x)$ into $g(x)$, as desired. $\blacksquare$