Question 556283
Determine the equation of the normal to the curve representing the function f (x) = 3sqrt (x), to the point where x = 9.
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{{{f(x) = 3x^(1/2)}}}
f'(x) = {{{3*(1/2)*x^(-1/2) = 3/(2sqrt(x))}}}
f'(9) = 1/2
f(9) = 9
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Slope m of normal = -2
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y = mx + b
9 = -2*9 + b
b = 27
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--> y = -2x + 27