Prove algebraically that the line $y=x+c$ intersects the curve $y=f(x)$ if $|a| \geq 1$ 
The function $f(x)$ is defined by $$f(x) = \frac{x(x-2)(x-a)}{x^2-1}$$
Prove algebraically that the line $y=x+c$ intersects the curve
$y=f(x)$ if $|a| \geq 1$, but there are values of $c$ for which there
are no points of intersection if $|a| < 1$.

Workings: I started off by letting $f(x) = x+c$ and then rerranging produces $$(a+c+2)x^2-(2a-1)x - c = 0$$
In order for the line $x+c$ and $f(x)$ intersect, then the discriminant $\Delta \geq 0 \iff (2a-1)^2-4(a+c+2)(-c) \geq 0$
Rerranging produces $$4a^2+4a+1+4ac+4c^2+8c \geq 0$$
This is the point where I start having difficulties. We have a discriminant in two variables. I notice that we can rewrite as a quadratic in $c$, so $$4c^2 + (8+4a)c + 4a^2+4a+1 \geq 0$$. If we want this quadratic in $c$ to be $\geq 0$ for all $c$ then this second discriminant must $\leq 0$. $$\Delta_2 = (8+4a)^2 - 16(4a^2+4a+1) \leq 0$$ Hence, $$\Delta_2 \leq 0 \iff 48-48a^2 \leq 0$$
So we see that when $|a| \geq 1$, the line intersects the curve.
However, I do not know how to approach the second part.
I.e.

but there are values of $c$ for which there are no points of
intersection if $|a| < 1$

Some additional questions, that I would like to ask.
Q1: Does the line $x+c$ intersect $f(x)$ for whatever $c$ value as long as $|a|\geq 1$. What if we viewed the first discriminant as a quadratic in $a$ instead of $c$, and got a range of values for $c$ for which the curves would not intersect? Would this mean that the curves would intersect only when $|a| \geq 1$ and not within some range of values of $c$
 A: 
I started off by letting $f(x) = x+c$ and then rerranging produces $$(a+c+2)x^2-(2a-1)x - c = 0$$

You have a typo here. It should be
$$(a+ c+ 2) x^2 - (2 a \color{red}{+} 1)x-c=0\tag1$$
I think that you have correctly showed that if $|a|\geqslant 1$, then $\Delta\geqslant 0$.
There is another way.
$$\begin{align}\Delta&=(-(2a+1))^2-4(a+c+2)(-c)
\\\\&=4c^2+(4a+8)c+(2a+1)^2
\\\\&=4\bigg(c^2+(a+2)c\bigg)+(2a+1)^2
\\\\&=4\bigg(\bigg(c+\frac{a+2}{2}\bigg)^2-\bigg(\frac{a+2}{2}\bigg)^2\bigg)+(2a+1)^2
\\\\&=4\bigg(c+\frac{a+2}{2}\bigg)^2-4\bigg(\frac{a+2}{2}\bigg)^2+(2a+1)^2
\\\\&=(2c+a+2)^2+3(a^2-1)\end{align}$$

*

*If $|a|\geqslant 1$, i.e. $a^2-1\geqslant 0$, then $\Delta\geqslant 0$.


*If $|a|\lt 1$, i.e. $a^2-1\lt 0$, then there are values of $c$ for which $\Delta\lt 0$ since $$\begin{align}\Delta\lt 0&\iff (2c+a+2)^2\lt 3(1-a^2)
\\\\&\iff -\sqrt{3(1-a^2)}\lt 2c+a+2\lt \sqrt{3(1-a^2)}
\\\\&\iff \frac{-a-2-\sqrt{3(1-a^2)}}{2}\lt c\lt \frac{-a-2+\sqrt{3(1-a^2)}}{2}\end{align}$$

Does the line $x+c$ intersect $f(x)$ for whatever $c$ value as long as $|a|\geq 1$

No, consider the following examples :

*

*If $a=1$, then $y=x-3$ does not intersect $y=f(x)$.


*If $a=-1$, then $y=x-1$ does not intersect $y=f(x)$.
So, in the first place, it is not true that the line $y=x+c$ intersects the curve $y=f(x)$ if $|a|\geqslant 1$.
(To notice the above examples, note that we have $x\not=\pm 1$ in the first place since the denominator of $f(x)$ is $x^2-1$. $(1)$ has a solution $x=1$ iff $a=1$. $(1)$ has a solution $x=-1$ iff $a=-1$. For $a=1$, $(1)$ can be written as $(x-1)((c+3)x+c)=0$. For $a=-1$, $(1)$ can be written as $(x+1)((c+1)x-c)=0$.)
So, we can say the followings :

*

*The line $y=x+c$ intersects $y=f(x)$ for whatever $c$ value except $c=-3,-1$ as long as $|a|\geqslant 1$.


*The line $y=x+c$ intersects $y=f(x)$ for whatever $c$ value as long as $|a|\color{red}{\gt} 1$.
