Determine the equation (s) of tangent (s) line to the curve y = 4x3 �

1.	Determine the equation (s) of tangent (s) line to the curve y = 4x3 – 3x + 5 which are perpendicular to the line 9y + x + 3 = 0.
Answer» finding the slope of the tangent by differentiating the curve \(\frac{dy}{dx}=12x^2-3\) m(tangent) \(=12x^2-3\) the slope of given line is \(-\frac{1}{9}\), so the slope of line perpendicular to it is 9 12x² -3 = 9 x = 1 or – 1 since this point lies on the curve, we can find y by substituting x y = 6 or 4 therefore, the equation of the tangent is given by equation of tangent is given by y – y₁ = m(tangent)(x – x₁) y – 6 = 9(x – 1) or y – 4 = 9(x + 1)

Determine the equation (s) of tangent (s) line to the curve y = 4x3 – 3x + 5 which are perpendicular to the line 9y + x + 3 = 0.

Answer»

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=12x^2-3\)

m(tangent) \(=12x^2-3\)

the slope of given line is \(-\frac{1}{9}\), so the slope of line perpendicular to it is 9

12x² -3 = 9

x = 1 or – 1

since this point lies on the curve, we can find y by substituting x

y = 6 or 4

therefore, the equation of the tangent is given by

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 6 = 9(x – 1)

y – 4 = 9(x + 1)