$(x-1)^{560}$ is said to have 560 roots. Now , all the roots must be 1(since it is the only factor tha can be formed ). And another example is $x^2 - 2x + 1 = 0 $ , Here $ x = 1$ ,only root as answer.
Q1 By saying $560$ roots , It does not give an difference to the answer( I mean that there is only root possible ) but does it give a difference in the graph of the function ?.
Roots of multiplicity become interesting when looking at the structure of a polynomial.
But, if you only care about the graph... if the multiplicity is even, the graph will touch the x-axis but not "cut through" the x-axis. If it is odd, it will cut through.
The graph will be "flat" in the neighborhood of a root of high multiplicity.
The graph of $y = (x-1)^{560}$ will be very nearly zero for all x in (0,2), and then quickly shoot up toward infinity for any x outside of this range.
The roots of these functions
$$f(x)=(x-1)^{560} \\ g(x)=x^2-4x+4=(x-2)^2$$
are not equal. This means that the graphs are also different.
I accept
$$f(x)=(x-1)^{560}, ~g(x)=(x-1)^2$$
Both functions get equal value at the point $x = 0,1,2.$ This means that the graphs pass through the same point, but it does not mean the graphs are the "same".
Just notice that:
This means, graps can never be the "same".
When will the graph of two functions be the same?
Of course, when these functions are equivalent functions.
For example:
$$\color{blue}{f(x)=|x|} \qquad \color{red}{g(x)=\sqrt {x^2}}$$
Since the values of the functions are the same at all points, the graphs are also the same.
