Number of inflection points of $(x-2)^6(x-3)^9$ 
Find the number of inflection points of $(x-2)^6(x-3)^9$

Attempt: 
If $f(x)$ has $n$ critical points then even $f(x+a)$ will also have $n$ critical points. 
So we can simplify it to finding the number of inflection points  of $(x)^6(x-1)^9$. 
I have evaluated the double derivative to be: 
$$6(x-1)^7x^4(35x^2-28x +5)$$
Clearly, the double derivative is $0$ at $4$ points, $0$, $1$ and the roots of the quadratic. But curvature is not changing around $x=0$ so it's not an inflection point. 
Thus, there should be $3$ inflection points. 
But answer given is $1$ inflection point only. 
Please let me know what concept am I missing on. 
I tried my level best to zoom into the graph and catch three critical points but there appears to be only one.  

It should be
$$f'(x)=3(x-3)^8(x-2)^5(5x-12)=0 \implies x=3,2,\frac{12}5$$
and
$$f''(x)=6(x-3)^7(x-2)^4(35x^2-168x+201)=0 \\\implies x_1=2,\,x_2=3,\,x_{3,4}=\frac{12}5\pm\frac{\sqrt{3/7}}{5}$$
and since $p(x)=35x^2-168x+201\implies p(12/5)\neq 0$ the unique stationary inflection point is at $x=3$ otherwise, if we look at general inflection points, by the sign of $f''(x)$ the inflection points are three at 
$$x_2=3,\,x_{3,4}=\frac{12}5\pm\frac{\sqrt{3/7}}{5}$$
