1

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 enter image description here

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?

Mick
  • 17,141
  • You have shown that if there are two fixed points, they must be $(-1, -1)$ and $(-2,-2)$. Now substitute these points and show they satisfy for any general $k$. – Macavity Mar 21 '14 at 02:04
  • Factorize the term as (x+1)(x+2)/k+1 +1, you can see that at - 1 and - 2, the value of y is independent of k. – Amandeephy Mar 21 '14 at 03:26
  • @Macavity Do you mean putting (-1, -1) or (-2, -2) back in to the original to see if the LHS constantly equal to RHS? (or to see that k disappears meaning it is independent?) – Mick Mar 21 '14 at 05:11
  • @A1D1S Do you mean after the (-1, -1) has been found, then factorize it as [(x+1)(x+2)/(k+1)] + 1? Even if it is so, (-1, -1) is still coming from some sort of verification - like using a particular case (like what I did by letting k = 0 first) to find the general case. – Mick Mar 21 '14 at 05:25
  • @Mick Yes. You need to show that for a general $k$ it works. – Macavity Mar 21 '14 at 10:41
  • No it is the other way around, see just open the k+7 term and split it into 2 one with 6x and one with ( k+1)x. For the above expression to always pass through a set of points it needs to be independent of k, which it would only be at - 1 and - 2 because the first term goes to 0 – Amandeephy Mar 21 '14 at 11:56
  • The original becomes $y = \frac 1 {k+1} [2x^2 + 6x + (k + 1)x + 4]$. What next? – Mick Mar 21 '14 at 17:15

0 Answers0