Linear Graph Intersection

If $a=1$ , then the three lines $y=f(x)$ , $y=g(x)$ , $y=h(x)$ are parallel (or coincide).

If $a \neq 1$ , then $f$ has a unique fixed point (the intersection of the lines $y=x$ and $y=f(x)$ ). Say this point is $t$ , that is $f(t)=t$ .

Now it's easy to see that $t$ is also a fixed point of $g$ and $h$ ; so all three lines pass through the point $(t,t)$ .

Since $y=g(x)$ and $y=h(x)$ are also straight lines, this point of intersection must be unique; ie $M,N,P$ are all the same point.

$f(x)$ = $ax + b$

$g(x)$ = $f(f(x))$

$h(x)$ = $f(f(f(x)))$

First, let's find the coordinate of point $M$ in general. At point $M$ , the $x$ value and the $y$ value of point $M$ satisfies both $f(x)$ and $g(x)$ . So, we can equalize both functions to find the value of $x$ that satisfies both functions.

$f(x)$ = $g(x)$

$ax + b$ = $f(ax + b)$

$ax + b$ = $a \times (ax + b) + b$

$ax$ = $a \times (ax + b)$

We divide both sides by $a$

$x$ = $ax + b$

$x - ax$ = $b$

$x \times (1 - a)$ = $b$

$x$ = $\frac{b}{1-a}$

Now substitute the $x$ value to any function to find the $y$ value. This time let's choose the function $f(x)$ .

$y$ = $ax + b$

$y$ = $a \times \frac{b}{1-a} + b$

$y$ = $\frac{ab}{1-a} + \frac{b \times (1-a)}{1-a}$

$y$ = $\frac{ab + b - ab}{1-a}$

$y$ = $\frac{b}{1-a}$

So, in general, point $M$ has the coordinate ( $\frac{b}{1-a}$ , $\frac{b}{1-a}$ ).

Next, let's find the coordinate of point $N$ in general. At point $N$ , the $x$ value and the $y$ value of point $N$ satisfies both $f(x)$ and $h(x)$ .

$f(x)$ = $h(x)$

$ax + b$ = $f(f(f(x)))$

$ax + b$ = $f(g(x))$ (because $g(x)$ = $f(f(x))$

$ax + b$ = $f( a \times (ax + b) + b)$

$ax + b$ = $a \times ( a^{2}x + ab + b ) + b$

$ax$ = $a \times (a^{2}x + ab + b)$

We divide both sides by $a$

$x$ = $a^{2}x + ab + b$

$x - a^{2}x$ = $ab + b$

$x \times (1 - a^2)$ = $b \times ( a + 1 )$

$x \times (1 + a)(1 - a)$ = $b \times (1 + a)$

$x \times (1 - a)$ = $b$

$x$ = $\frac{b}{1-a}$

Now substitute the $x$ value to any functions to find the $y$ value. This time let's choose the function $h(x)$ .

$y$ = $a \times ( a^{2}x + ab + b ) + b$

$y$ = $a^{3}x + a^{2}b + ab + b$

$y$ = $a^{3} \times \frac{b}{1-a} + a^{2}b + ab + b$

$y$ = $\frac{a^{3}b}{1-a} + \frac{ a^{2}b \times (1-a) }{1-a} + \frac{ ab \times (1-a) }{1-a} + \frac{b \times (1-a) }{1-a}$

$y$ = $\frac{ a^{3}b + a^{2}b - a^{3}b + ab - a^{2}b + b - ab }{1-a}$

$y$ = $\frac{b}{1-a}$

So, in general, the point $N$ has the coordinate ( $\frac{b}{1-a}$ , $\frac{b}{1-a}$ ).

Now that the intersection point of ' $f(x)$ and $g(x)$ ' and ' $f(x)$ and $h(x)$ )' is the same, it can't make a triangle since there are two points that coincide. because $M$ and $N$ are same, then $P$ must also be the same point, because if $P$ is not the same point, then the lines will intersect at three different points. Therefore, $M$ , $N$ , and $P$ are the same point.