Question

Find the equations of the tangent and normal lines to the curvey y = 3x² - 2x + 1 at the point (1,2).​

Answer 1

Answer:

slope of tangent y=3x^2-2x+1

dy/dx =6x-2

(dy/dx) at x=1 and y=2

dy/dx=6*1-2

=4

thus equation of tangent have slope 4 and Passing points 1,2 is given by

y-y1=m(x-x1) where m is slope

y-3=4(x-1)

y-3=4x-4

y-3+4=4x

y+1=4x

thus equation of line is 4x-y-1 ...

hope it helps

have a nice day....