Question

Find the equation of the tangent line, y = x^2 + 4x - 1 at x = 2

Answer 1

65

Step-by-step explanation:

3x - y + 1 =0

d/dx (3x) - dy/dx + d/dx (1) = 0

dy/dx = 3

y = x2 + 4x - 16

dy/dx = 2x + 4

Hence

2x + 4 = 3

x= 3-4/2 = -1/2

at x = -1/2y = (-1/2)2 + 4 (-1/2) - 16 = 1/4 -2 -16

y = -71/4

so the point p (-1/2, -71/4)

equation of tangent

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

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

3x - y = 71/4 - 3/2 = 71-6/4 = 65/4

12x - 4y = 65.

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