Question 1114990
 Find the equation of the straight line perpendicular to line y =x^ 2+ 1,and
 passing through the point (4,5).
slope of y = x^2+1 at every point = y' = 2x
slope of curve at x = 4 is 2*4 = 8
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slope of line perpendicular to the curve at (4,5) is -1/8
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Form:: y = mx + b
5 = (-1/8)4+b
b = 5+1/2 = 11/2
Equation:
y = (-1/8)x+(11/2)
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What is the gradient of y = x^3/2,when x = 4? 
slope at every value of "x" is (3/2)x^(1/2)
gradient = slope at x = 4 is (3/2)4^(1/2) = 3
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Cheers,
Stan H.
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