equation of tangent line to a graph at a point using derivat
- Thread starter twiggyg93
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skeeter
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- Dec 15, 2005
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you need two things ...
the coordinates (x,y) of the point of tangency, and
the value of the slope at that point.
you've been given the x-value for the point of tangency, how will you find the y-value?
what expression gives you the slope at any point on the given curve? how would you use that expression to find the specific slope?
once you've found those two things, use the point-slope form of a linear equation to get the desired equation.
the coordinates (x,y) of the point of tangency, and
the value of the slope at that point.
you've been given the x-value for the point of tangency, how will you find the y-value?
what expression gives you the slope at any point on the given curve? how would you use that expression to find the specific slope?
once you've found those two things, use the point-slope form of a linear equation to get the desired equation.