How much progress have you made with this question?

Find the slope of y= x^2 +6x +10 at (-2,2):

y' = 2x + 1 so slope is -3

The normal line will have slope 1/3 (negative reciprocal).

So the slope to ax^2 + bx +? 2 at (1, 1/2) is 1/3.

Plug in the point (1, 1/2) into the function to get one condition on a and b.

Plug in the point x=1 and m=1/3 into the derivative of the curve.

Then solve for a and b.

Good luck!!