I have some doubts in proving the following:-
C is the curve $$y = \frac 1 {k+1} [2x^2 + (k + 7)x + 4]$$, where k is a real number not equal to -1. Show that C always passes through two fixed points for all allowable values of k. (What are the co-ordinates of these two points?)
Here is what I have done:-
(0) using k = 0, 1, –2 to check from Wolfram-alpha and get
In order to get those two points (of intersection of two different C’s),
(1) I let m and n be two different allowable values of k;
(2) Set up the following two equations and solve; $$Y = \frac 1 {m + 1} [2x^2 + (m + 7)x + 4] …… [@]$$ $$Y = \frac 1 {n + 1} [2x^2 + (n + 7)x + 4] …… [!]$$ (3) Performing (m + 1) * [@] – (n +1) * [!], I arrive at y = x.
(4) Since it is true for all values of k except –1, I let k = 0 and get $x = 2x^2 + 7x + 4$.
(5) From that, I get (–1, –1) and (–2, –2).
Here is my question "is the proof valid?".
I raise this question because:-
(1) At the most, I just have y = x correct. There are so many points satisfying that (not just two).
(2) The rest is just checking, verifying and assuming the given but not showing it at all.
Any idea?