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Find an equation of the tangent line to the curve at the given point.
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\[y= \sqrt{x} (9,3)\]
find the derivative first! know how to ?
f(x)-f(a)/x-a
derivative of y at point 9,3 will give you slope of tangent at that point
I came up with \[-\frac{ 1}{ 2 }\]
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but that's the wrong answer
you mean the slope as -1/2 ?
i mean like y= \[-\frac{ 1 }{ 2 }\]
ofcourse thats not correct...the slope isn't 0 have u solved a similar question before ?
no
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ok, so you need to use the definition of derivative, right ? "f(x)-f(a)/x-a"
yes thats what i should use but i don't know where to put what
|dw:1411783189182:dw| you can solve limits question ?
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