Find the equation of the tangent to the parabola `y=x^2-2x+3` at point (2, 3).

Explanation

Since the equation of the parabola is not in the standard from, we use calculus method to find the equation of tangent.
`y=x^(2)-2x+3`
Differentiating w.r.t. x, we have
`(dy)/(dx)=2x-2`
We want to find the tangent at point (2,3). Then
`((dy)/(dx))_(2,3)=2(2)-2=2`
Hence, using point-slope form, the equation of tangent is
`y-3=2(x-2)ory=2x-1`