Question

1. Find the equation of a normal to the curve y= 2² - 2x +3 at the point (3,0)​

Answer 1

I think you meant to say the equation is

y = 2x² - 2x + 3

Differentiate both sides with respect to x :

dy/dx = 4x - 2

At the point (3, 0), the slope of the tangent line is dy/dx(3) = 4•3 - 2 = 10. Then the normal line to the curve at (3, 0) has slope -1/10.

Using the point-slope formula, the equation of the normal line is

y - 0 = -1/10 (x - 3)   ⇒   y = (3 - x)/10