Answer:
The correct answer is option (C), that is, [tex]f(x)=x^{4}(x-a)(x-b)^{2}[/tex]
Step-by-step explanation:
From the given graph, we can see that the left most point where graph meets axis is the point where graph touches the x axis. Therefore, the factor corresponding to that point must have an even multiplicity. Therefore, exponent of one of (x-a) or (x-b) must be even.
Further, at x=0, the graph touches x axis. Therefore, multiplicity of this zero must be even as well. This implies that exponent of x must be even.
Finally, at right most zero, we can see that the graph cross x axis, without staying on x axis. Therefore, multiplicity of that particular zero must be odd and more accurately it should be 1.
Combining all these factors together, we can conclude that one of (x-a) and (x-b) must have an exponent 1 and other must have an even exponent. At the same time x must have an even exponent. All these conditions are satisfied by the function [tex]f(x)=x^{4}(x-a)(x-b)^{2}[/tex]. Here, x has an even exponent (4). (x-a) has exponent 1 and (x-b) has an even exponent (2). Therefore, correct answer is option (c).
