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.  
 A: 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 
$$x_2=3,\,x_{3,4}=\frac{12}5\pm\frac{\sqrt{3/7}}{5}$$
A: It may help to see a very vertically magnified plot of the second derivative (of the original untranslated function).

Given that the inflection points can only be at $x = 2$, $x = 3$, or the roots of the quadratic term $35x^2-168x+201$ (i.e., $r_{1, 2} = \frac{12}{5} \pm \frac{\sqrt{21}}{35}$), the plot above shows that there must be three inflection points: at $x = r_1 \approx 2.269$, $x = r_2 \approx 2.531$, and $x = 3$.
A: It should be obvious from the degree of the factors:
$f(2) = 0, f(3) = 0\\
f'(2) = 0, f'(3)= 0\\
f''(2) = 0, f''(3)= 0$
Since $f(x) \le 0$ when $x< 3$ and $f(x) > 0$ when $x>3,$ $f(2)$ is a maximum (and not an inflection point, and $f(3)$ is an inflection point and not a min or a max.
There must be a point $c\in (2,3)$ where $f'(c) = 0$ and it must be a local minimum.
There is a point of inflection in $(2,c)$ and one in $(c,3)$
That is 3 points of inflection.
