How do you find the equation of a line normal to the function #y=5/sqrtx-sqrtx# at (1,4)?

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Answer 1 · 2017-05-18T18:43:42

The equation is #y = 7 - 3x#.

#y = 5x^(-1/2) - x^(1/2)#

This has derivative

#y' = -5/2x^(-3/2) - 1/2x^(-1/2)#

#y' = -5/(2x^(3/2)) - 1/(2x^(1/2)#

Hence, the slope of the tangent is:

#y'(1) = -5/(2(1)^(3/2)) - 1/(2(1))^(1/2)#

#y'(1) = -5/2 - 1/2#

#y'(1) = -3#

We now have the equation as:

#y- y_1 = m(x - x_1)#

#y -4 = -3(x - 1)#

#y - 4 = -3x + 3#

#y = 7 - 3x#

Hopefully this helps!