Question

find all values of x for which the tangent to y=(x+2)^2 passes through the origin

Answer 1

The slope at any point is dy/dx

dy/dx = 2x + 4

Equate this slope with the slope between the origin (0,0) and any given point (x,y)

slope = (y2-y1)/(x2-x1) = y-0/x-0 = (x+2)^2/x

--> (x+2)^2/x = 2x+4
(x+2)^2 = x(2x+4)
x^2 +4x +4 = 2x^2 +4x
x^2 = 4
x = +-2

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