Question

Show that the tangent to the curve y = x³ – 6x² + x + 3 at the point (0, 3) is parallel to the line y = x + 5.

Answer 1

y = x³ – 6x² + x + 3

∴ `dy/dx = d/dx (x^3 - 6x^2 + x + 3)`

= 3x² – 6 × 2x + 1 + 0

= 3x² – 12x + 1 

∴ `(dy/dx)_("at" (0","  3)) = 3(0) - 12(0) + 1 = 1`

= Slope of the tangent at (0, 3)

Also, slope of the line y = x + 5 is 1.

∵ Slope of the tangent at (0, 3) = Slope of the line y = x + 5.

Hence, the tangent to the curve at (0, 3) is parallel to the line y = x + 5.