Given $f(x) = x^2$ - NSC Mathematics - Question 9 - 2022 - Paper 1

Expanding the distance formula:

$d = ext{sqrt}((x - 10)^2 + (x^2 - 2)^2)$

This simplifies to:

$d = ext{sqrt}((x - 10)^2 + (x^4 - 4x^2 + 4))$

Continuing, we can combine the terms:

$= ext{sqrt}(x^4 - 4x^2 + (x - 10)^2 + 4)$ $= ext{sqrt}(x^4 - 4x^2 + x^2 - 20x + 100 + 4)$

Next, find the derivative of the squared distance to minimize the distance:

$d^2 = x^4 - 5x^2 - 20x + 104$

Taking the derivative:

$\frac{d}{dx}(d^2) = 4x^3 - 10x - 20$

Setting the derivative equal to zero gives:

$4x^3 - 10x - 20 = 0$

We can attempt to find the critical points by solving:

$4x^3 - 10x - 20 = 0$

Testing likely candidates, we can observe:

Trying $x = 2$:

$4(2)^3 - 10(2) - 20 = 0$

Thus, $x = 2$ is a critical point.

Now, substitute $x = 2$ back into the distance formula:

$d = ext{sqrt}((2 - 10)^2 + (2^2 - 2)^2)$ $= ext{sqrt}((-8)^2 + (0)^2)$ $= ext{sqrt}(64)$ $= 8$

Therefore, the minimum distance from the point $(10, 2)$ to the curve $f$ is 8.