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December 20, 2014

December 20, 2014

Posted by **nick** on Sunday, September 30, 2007 at 7:01pm.

f(x) = x^3 - x^2 - 4x + 4

The point (a,b) is on the graph of f and the line tangent to the graph at (a,b) passes thru the point (0, -8) which is not on the graph of f. Find the values of a and b.

I have no clue how to solve this

I know the derivative of f prime (x) is

3x^2 - 2x - 4 and so (a,b) would be on the derivative line and so would (0, -8) but how do i find (a,b)???

- calc -
**drwls**, Sunday, September 30, 2007 at 7:23pmThe equation of the tangent line is

y = mx -8,

where m is the slope at the tangent point (a,b).

At (a,b), the value of y is

a^3 - a^2 + 4a + 4 = b

and the slope is

dy/dx @ x=a =

3x^2 -2x -4 = 3a^2 -2a -4 = m

From the tangent line equation, we also know that, at (a,b)

b = m a -8

So there are 3 equations in the three unknowns, m, a and b.

It is a bit messy, but that is as far as I can go.

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