I don't get what equation to use.
What point on the graph of y=x^2-2x+1 is closest to the point (1,2)?
Is this problem from a calculus class? Because I think that may be the only way to solve it, and it's a kinda long solution (and I left out a few things)... unless you just graph it and try to estimate.
It's a minimization problem. You need to use the distance formula
D = d2 = (x2 - x1)2 + (y2 - y1)2
to find the distance between (1, 2) and some point on the graph. The closest point to (1, 2) will be where the D (for convenience, i let d2 = D, since they're both a minimum at the same point) is at a minimum. Let (x1, y1) be (x, y), and (x2, y2) be (1, 2):
D = (x - 1)2 + (y - 2)2
If we differentiate this w.r.t x, the minimum distance will be where dD/dx = 0. ie we need to minimize D. I'll use implicit differentiation. The second line is just distributing.
dD/dx = 2(x - 1) + 2(y - 2)(y')
dD/dx = 2(x - 1) + 2yy' - 4y'
Now, set dD/dx = 0, and substitute in y (factorize y first),
y = x2 - 2x + 1 = (x - 1)(x - 1)
so dy/dx = 2x - 2 = 2(x - 1)
dD/dx = 0 = 2(x - 1) + 2(x - 1)(x - 1)*2(x - 1) - 4*2(x - 1)
Notice we have some common factors - we can divide both sides by 2(x - 1), to get:
0 = 1 + 2(x - 1)(x - 1) - 4
Now we just have to solve that for x, so distribute/FOIL the parentheses
0 = 1 + 2x2 - 4x + 2 - 4
0 = 2x2 - 4x - 1
This doesn't factor, so use the quadratic formula to find that x = 1 + (√6)/2 and x = 1 - (√6)/2
Those are our two x-coordinates, then you plug one of those into y = x2 - 2x + 1 to find the y-coordinate.