Question

find the points on the curve given by y=x^3-6x^2+x+3 where the tangent are parallel to the line y=x+5

Answer 1

answer

(0,3),(4,-25)

step by step explanation

given curve y=x³-6x²+x+3

slope dy/dx=3x²-12x+1

given tangent is parallel to y=x +5

therefore from above line

dy/dx=1

3x²-12x+1=1

3x²-12x=0

3x(x-4)=0

x=0,4

consider x=0 and substitute in the given curve

y=3

now consider x=4 and substitute in the given curve

y=(-25)

therefore the points are (0,3),(4,-25)