I don't get what equation to use.

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 = d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{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 d^{2} = D, since they're both a minimum at the same point) is at a minimum. Let (x_{1}, y_{1}) be (x, y), and (x_{2}, y_{2}) 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** = x^{2} - 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 + 2x^{2} - 4x + 2 - 4

0 = 2x^{2} - 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 = x^{2} - 2x + 1 to find the y-coordinate.

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