Since (a, b) is on the graph of y = f(x), we can say that: f(a) = b
Let's plug in (a-6)/3 for x into the equation for the second graph.
y = -3 * f( 3 [ (a-6)/3 ] + 6) -2 = -3 * f(a) - 2 = -3b - 2
When x = (a-6)/3 , y = -3b - 2 so we can say that the point \((\frac{a-6}{3}, -3b-2)\) must be on the second graph.
Check: https://www.desmos.com/calculator/aqfexnjff4
(On that graph we can move a around and also change the definition of f , and for each the cases I tried this seems to hold)
\(3\lt\sqrt{2x}\lt4\\~\\ 3^2\lt(\sqrt{2x}\ )^2\lt4^2\\~\\ 9\lt2x\lt16\\~\\ \frac92\lt\frac{2x}{2}\lt\frac{16}{2}\\~\\ 4.5 \lt x \lt 8\)
The integers between 4.5 and 8 exclusively are:
5, 6, 7
So there are 3 different integer values of x that satisfy the condition.
Check: https://www.wolframalpha.com/input/?i=3%3Csqrt%282x%29%3C4