At the point \(x=1\), the derivative is \(f'(1)=2(1)-2=0\). This means that the tangent line to the graph of \(f(x)\) at the point \((1,f(1))\) is horizontal.
To find the equation of the tangent line, we use the point-slope form:
$$y-y_1=m(x-x_1)$$
where \((x_1,y_1)\) is the point on the tangent line that we know, and \(m\) is the slope of the tangent line.
In this case, \((x_1,y_1)=(1,f(1))=(1,2)\) and \(m=f'(1)=0\). So the equation of the tangent line is:
$$y-2=0(x-1)$$
Simplifying, we get:
$$y=2$$
So the tangent line to the graph of \(f(x)=x^2-2x+3\) at the point \((1,2)\) is the horizontal line \(y=2\).