Equation of a tangent line give a point and slope

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November 9th 2009, 07:49 AM
shelbilynn08

Equation of a tangent line give a point and slope

Im trying to write an equation A(w) for the area of the triangle formed by the coordinate axes and the line tangent to the graph of f(x)=6-x^2 at the point (w,6-w^2), given that the slope of the line is -2w.

Anyone have any clue on where to even start?!
November 9th 2009, 08:22 AM
skeeter

Quote:

Originally Posted by shelbilynn08

Im trying to write an equation A(w) for the area of the triangle formed by the coordinate axes and the line tangent to the graph of f(x)=6-x^2 at the point (w,6-w^2), given that the slope of the line is -2w.

Anyone have any clue on where to even start?!

Area = (1/2)(x-intercept of the tangent line)(y-intercept of the tangent line)

tangent line equation ...

$y - (6-w^2) = -2w(x - w)$

x-intercept, y = 0 ...

$0 - (6-w^2) = -2w(x - w)$ , solve for x in terms of w

y-intercept, x = 0 ...

$y - (6-w^2) = -2w(0 - w)$ , solve for y in terms of w
November 10th 2009, 09:34 AM
shelbilynn08

thanks for you help but how would i explain how i got that answer.. my teacher is strict about decribing it and i honestly dont even understand how to do what you told me hah