How many horizontal tangents does the graph of f(x)=(x-1)^2(x-3)^2 have?

f(x)=(x-1)^2(x-3)^2

=2(x-1)(1)+2(x-3)(1)
=2x-2+2x-6

f'(x)=4x-8

0=4x-8
=4(x-2)

x=2

Therefore it only has 1 horizontal tangents?

To find the number of horizontal tangents of a graph, we need to determine the number of points where the derivative of the function, f'(x), equals zero.

In this case, the derivative of f(x) is given by f'(x) = 4x - 8. To find the points where f'(x) = 0, we set 4x - 8 equal to zero and solve for x:

4x - 8 = 0
4x = 8
x = 2

So, we have found that x = 2 is the only value of x where the derivative equals zero, which means it is the only point where the graph of f(x) has a horizontal tangent line.

Therefore, the graph of f(x) = (x - 1)^2(x - 3)^2 has one horizontal tangent.