The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all real values for $x\in R$ are: => The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$

takes all real values for $x\in R$ are:

My question is why we need to validate end points i.e. $1,7$ (Refer the last part of my attempt)
My attempt is as follows:-
$$y=\dfrac{ax^2+3x-4}{a+3x-4x^2}$$
$$ya+3yx-4yx^2=ax^2+3x-4$$
$$x^2(-4y-a)+x(3y-3)+ya+4=0$$
As $x$ can be any real, so $D\ge0$
$$9y^2+9-18y-4(ya+4)(-4y-a)\ge0$$
$$9y^2+9-18y+4(4y^2a+ya^2+16y+4a)\ge0$$
$$y^2(9+16a)+y(4a^2+64-18)+9+16a\ge0$$
$$y^2(9+16a)+y(4a^2+46)+9+16a\ge0$$
As range is $R$, so discriminant of quadratic in $y$ should be less than equal to zero
$$4(2a^2+23)^2-4(9+16a)^2\le0$$
$$(2a^2+23-9-16a)(2a^2+23+9+16a)\le0$$
$$(2a^2-16a+14)(2a^2+16a+32)\le0$$
$$(a^2-8a+7)(a^2+8a+16)\le0$$
$$a\in[1,7]$$
But in such type of questions, we always check at endpoints like here we need to check at $a=1$ and $a=7$. But I don't understand what is so special about endpoints.
From the above calculation I can only say at $a=1,7$ discriminant of quadratic in $y$ is zero, but what is so special about this. Please help me in this.
 A: When $a=1$, $\displaystyle y=\frac{x^2+3x-4}{1+3x-4x^2}=\frac{-x-4}{1+4x}\ne-\dfrac14$ for all $x\in\mathbb{R}$.
When $a=7$, $\displaystyle y=\frac{7x^2+3x-4}{7+3x-4x^2}=\frac{7x-4}{7-4x}\ne-\dfrac74$ for all $x\in\mathbb{R}$.
A: The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all real values for $x\in R$
$\dfrac{ax^2+3x-4}{a+3x-4x^2}$ will take all the values for $x \in \mathbb R$ only if the denominator is not equal to zero!
So, For $a + 3x - 4x^2 \neq 0; \iff -3 \pm \sqrt{9 +16a} \neq 0$
We need to check all the points for which the denominator will become equal to zero, and we need to remove those values of $a$ which includes $1, 7$.
A: Let's denote $f(x)=ax^2+3x−4$, $g(x)=-4x^2+3x+a$, and $h(x)=f(x)/g(x)$.
Since $a>0>-9/16>-4$, $f(x)$ and $g(x)$ cannot have the same roots, and each of them has two distinct roots. However, they may have a common root. If $a=1,7$ then there exists $y_0\in \mathbb{\mathbb{R}}$ for which only one $x$ may give $h(x)=y_0$. This $x$ may be the root of the denominator, so we must check. Else, when $1<a<7$, there exists two distinct candidates $x_1,x_2$ which may give $h(x)=y$. In this case at least one of $x_1,x_2$ is not a root for $g$, since otherwise they will be the roots of $f$.
