Use Newton's Method to approximate $x$, accurate to within $10^{-4}$ that produces the point on $y=x^2$ closest to $(1,0)$ 
Use Newton's Method to approximate $x$, accurate to within $10^{-4}$ that produces the point on $y=x^2$ closest to $(1,0)$.

$\textbf{My approach:}$
I consider the distance between some arbitrary point on $y=x^2$ and $(1,0)$, this is given by:
$$\begin{align}d(x) &= \sqrt{(x-1)^2 + (x^2-0)^2} \\ &= \sqrt{x^4 + x^2 -2x +1} \end{align}$$
Now, if $d$ is to have a minimum at some point $x_0$, it's slope must be $0$ there, so I want to use Newton's Method to approximate $x_0$ such that $d'(x_0)=0$. In addition, since we require a minimum, we also want $d''(x_0)>0$.
$\textbf{Some questions:}$
Using a program to apply Newton's Method in computing $d'(x)=0$, after 200 iterations with initial value $0.5$, I am nowhere close to the expected root $0.5897$. Is my problem that of finding a good initial value?
I noticed that computing $d''(x)=0$ gives me the result after 3 iterations. Why does finding the root of the second derivative of $d$ give the result when my problem is to minimize $d$?
 A: You can define your objective function as
$h(x) = x^4+x^2-2\,x+1$
$h'(x) = 4\,x^3+2\,x-2$
$h''(x) = 12\,x^2+2$
We can see that $h''$ is strictly positive, so you do not have to worry about it, you only need to find the point where $h'=0$.
So, using Newton's Method:
$x_0=0.5\\
x_1 = x_0-\frac{h'(x_0)}{h''(x_0)} = 0.5 - \frac{0.5+1-2}{3+2} = 0.6\\
x_2 = x_1-\frac{h'(x_1)}{h''(x_1)} = 0.6 - \frac{0.864+1.2-2}{4.32+2} = 0.5899$
Since you are not getting the right result, probably there is an implementation error.
A: You need to start close to the solution.
With the objective $d$, I get
$x_0 = 0.5, x_1\approx 0.6087, x_2\approx 1, x_3 \approx 0.5898$.
If you start at $x_0 = 2$, for example, the iterates go 'wild'.
The function $d^2$ is a little better behaved from a Newton's method perspective as Daniel's answer illustrates.
A: For $y = x^2$,
the slope at
$(a, a^2)$
is $2a$,
so the slope of the normal is
$-\frac1{2a}$.
The equation of the normal is thus
$\frac{y-a^2}{x-a}
=-\frac1{2a}
$.
If this passes through
$(1, 0)$,
then
$\frac{-a^2}{1-a}
=-\frac1{2a}
$
so that
$2a^3 = 1-a$
or
$2a^3+a-1 = 0$.
According to Wolfy,
this has a real root of about
$0.589754512301458$
and two complex roots of about
$0.294877256150729 \pm 0.872271625461329 i
$.
This is easily generalized.
