Hint: Minimizing the distance is equivalent to minimizing the square of the distance. Remove that square root sign. You get a harmless quadratic.
So I would write "equivalently, we minimize the square of the distance $\dots$. "
The calculation can be done without removing the square root sign, but the probability of mechanical error increases markedly.
$$ d = \sqrt{(x - (-2))^2 + ((3x^2+4x-8) - (-3))^2} = \sqrt{\left(x+2\right) ^2 + \left(3x^2+4x-5 \right)^2}$$
Instead of minimizing $d$, lets minimize $d^2$. It should be noted that the minimum of $d$ and $d^2$ are exactly the same as $d \geq 0$. This gets rid of the square root, thereby simplifying the problem.
$$d^2 = (x+2)^2 + (3x^2+4x-5)^2.$$
Now, we can use calculus to minimize $d^2$,
$$(d^2)' = 2(x+2) + 2(3x^2+4x-5)(6x+4) = 36x^3 + 72x^2 - 26x - 36$$
This has roots at $x \approx -2.118, -.631, .749$. We know that one of these must be the absolute minimum, so evaluate $d^2$ for each $x$-value and whichever is smallest is the $x$-value of the point on the parabola nearest to the given point. You should then plug that $x$-value into the equation for the parabola to determine the $y$-value where this occurs. I got,
Ans: $ (-2.118, -3.014) $
Best Answer
Hint: closest point lies on the line, connecting the center of the circle and this point. Proven by triangle inequality.