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y=sqrtx/(x+7) find equation of the tangent line at (9,3/16) and the equation of the normal line at (9,3/16)
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What can be said about the derivative of a curve at a point with its tangent line?
not sure
@zepdrix
So we can write the tangent line as,\[\Large y-y_o\quad=\quad m(x-x_o)\] Where our given point is \(\Large (x_o,\;y_o)\quad=\quad (9,\; 3/16)\) And where \(\Large m\) is given by derivative function evaluated at x=9. \(\Large f'(9)=m\)
Have you tried taking the derivative of your function yet?
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yes i got (7-x)/2sqrtx(x+7)^2
i was having trouble with the normal line equation. i dont get what that means
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