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Ricardo S

Guest

Tangent Line Equation Please Help

What is the equation of any tangent to the graph y=x^3 that passes through the point (0,-16)?

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Tangent Line Equation Please Help

First, the tangent intersects in the positive region, due to the convex curvature there as seen from (0,-16).  Going down and left, it is concave so won't work.
Then the slope must be the derivative.
So (y+16)/x = d(x^3)/dx and y = x^3 must both be true at same time.
(y+16)/x = 3x^2 and y = x^3
y+16 = 3x^3 and y = x^3
y = 3x^3 - 16 = x^3
2x^3 = 16
x^3 = 8
x = 2
y = 8
tangent goes through (2,8) at intersection point.
slope is (8+16)/2 --> 12
y - y1 = m(x-x1)
y - 8 = 12(x - 2)
y - 8 = 12x - 24
y = 12x - 16
or y = 4(3x - 4)
Please check my answer.

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