I understand how to find the equation of a tangent and a normal line at a given point by using the first derivative, and I know how to find the distance from x,y to 2,1 using the distance formula, but I don't see how to put it together. The book says:

Use the distance formula [D(x)]^2 = [(x-2)^2] + [(5+3x-2x^2) - 1]^2
and set D'(x) = 0 which is (4x^3) - (9x^2) - 3x + 5 = 0
and solve the resulting polynomial equation using Newton’s method.

I can get [(x-2)^2] + [(5+3x-2x^2) - 1]^2 =
[x^2+4x+4] + [4x^4 - 12x^3 - 7x^2 + 24x + 16] =
[4x^4 - 12x^3 - 6x^2 + 20x + 20] =
4[x^4 - 3x^3 - (6/4)x^2 + 5x + 5]
and I can get the first derivative
4x^3 - 9x^2 - 3x = 5
BUT I DON'T SEE WHY WE AREN'T TAKING THE DERIVATIVE OF
SQRT(4[x^4 - 3x^3 - (6/4)x^2 + 5x + 5] )

AND FROM HERE I'M LOST. EVEN IF THE BOOK'S DERIVATIVE IS RIGHT, I DON'T
KNOW WHAT TO DO WITH IT.

We want to find the point on the curve y=-2x^{2}+3x+5 that is closest to (2,1).

In problems of minimizing or maximizing a distance, there is a trick that is helpful for simplifying computations. It is based on the observation that the distance and the square of the distance have their max and min at the same point. That is why we do not use the square root. Makes things easier anytime you can eliminate a radical.

L=\sqrt{(x-2)^{2}+(y-1)^{2}}

L=\sqrt{(x-2)^{2}+(-2x^{2}+3x+4)^{2}}... [1]

So, from the observation mentioned above, we can use:

S=L^{2}=(x-2)^{2}+(-2x^{2}+3x+4)^{2}

S=L^{2}=4x^{4}-12x^{3}-6x^{2}+20x+20

Differentiate and get \frac{dS}{dx}=16x^{3}-36x^{2}-12x+20

Set to 0 and solve for x.

You will get 3 solutions. You can use the 2nd derivative test to check the nature of these critical point.

Better yet, just plug your solutions into [1] to see which gives the shortest distance. You can also graph it and see.

2x+2-1-2