Answer
The function could be $f(x)=ax(x-1)^2(x-2), \text{ where }a>0$
Work Step by Step
Thanks to the real zeros we see on the graph, we can find the function through its factored form:
$f(x)=a(x-0)(x-1)^2(x-2)$
That's because when any (x-r) equals zero, f(x) equals zero. Also, since the graph touches the x-axis at x=1, the factor (x-1) is raised by an even exponent.
The only thing left is to find 'a'. Nonetheless, since there are no non-x-intercept points given to solve the equation for 'a', we can leave the answer as:
$f(x)=ax(x-1)^2(x-2), \text{ where }a>0$
The reason 'a' must be a positive number is that the end behavior of the graph resembles $y=x^4$. If 'a' were to be negative, the graph will look flipped like $y=-(x^4)$.