SOLUTION: find where the tangent to the curve y=x^3 at the point where x=2, meets the curve again

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Question 1192494: find where the tangent to the curve y=x^3 at the point where x=2, meets the curve again
Answer by greenestamps(13195) About Me  (Show Source):
You can put this solution on YOUR website!


f(x) = x^3

f(2) = 8

f'(x) = 3x^2

f'(2) = 12

The tangent has slope 12 and passes through the point (2,8); its equation is y=12x-16.

We are to find the other value of x where the graphs of x^3 and 12x-16 intersect.

x%5E3+=+12x-16
x%5E3-12x%2B16=0

We could use standard methods to try to find how that cubic expression factors. However, we know something about that expression that makes it easy to factor.

We know that the graph of y=12x-16 is tangent to the graph of y=x^3 at x=2. That means that cubic equation has a double root at x=2. Removing that double root shows us that

x%5E3-12x%2B16=%28x%2B4%29%28x-2%29%5E2

so the other point where the tangent intersects the graph of y=x^3 is when x=-4; that point of intersection is (-4,-64).

A graph....

graph%28400%2C400%2C-5%2C5%2C-100%2C60%2Cx%5E3%2C12x-16%29