Find the equation of the tangent line to the curve y = x2 + 4x – 16

1.	Find the equation of the tangent line to the curve y = x2 + 4x – 16 which is parallel to the line 3x – y + 1 = 0.
Answer» Given as y = x² + 4x - 16 Differentiate with respect to x, to get the slope of tangent dy/dx = 2x + 4 m(tangent) = 2x + 4 The equation of tangent is given by y – y₁ = m(tangent)(x – x₁) Comparing the slope of tangent with given equation 2x + 4 = 3 x = -1/2 Substitute the value of x in the curve to find y y = (1/4) - 2 - 16 = - 71/4 So, the equation of tangent is parallel to the given line is y + (71/4) = 3(x + (1/2))

Find the equation of the tangent line to the curve y = x2 + 4x – 16 which is parallel to the line 3x – y + 1 = 0.

Answer»

Given as y = x² + 4x - 16

Differentiate with respect to x, to get the slope of tangent

dy/dx = 2x + 4

m(tangent) = 2x + 4

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

Comparing the slope of tangent with given equation

2x + 4 = 3

x = -1/2

Substitute the value of x in the curve to find y

y = (1/4) - 2 - 16 = - 71/4

So, the equation of tangent is parallel to the given line is

y + (71/4) = 3(x + (1/2))