find the equation of the tangent and normal line to the curve y = (3 - x) ¹² at the point ( 2:1)

Question

Please answer the following question on calcul

Answer 1

dy/dx=-12(3-x)¹¹=-12(3-2)¹¹=-12 at (2,1) is the gradient of the tangent and 1/12 the gradient of the normal.

The equation of the tangent line is y-1=-12(x-2), y=-12x+25;

The equation of the normal line is y-1=(1/12)(x-2), 12y-12=x-2, so y=(1/12)(x+10).

Answer 2

y=(3-x)¹²    @(2,1)

y'= 12(3-x)¹¹ (1)

y'= 12(3-x)¹¹ 

y'= 12(3-2)¹¹ 

y'= 12 

M= 12

Y-Y₁=M(X-X₁)

Y-1=12(X-2)

Y-1=12X-24

Y-1-12X+21=0

Y-12X+23=O   -------> TANGENT LINE 

M= 12

Y-Y₁=M(X-X₁)

Y-1=-1/12 (X-2)

12[Y-1=-1/12 (X-2)]12

12Y-12=-X+2

12Y-12+X-2=0

12Y+X-14=0 --------> NORMAL LINE