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Find the slope of the normal line to y= ln(15-x) at x=4
14 years ago

Answers

Reiny
when x=4, y = ln(11)
so our point is (4,ln11)

dy/dx = -1/(15-x)
which, when x=4 is -1/11

so the slope of the normal is +11

equation:
y - ln11 = 11(x-4)
11x - 44 = y-ln11
11x - y = 44 - ln11 OR y = 11x + ln11 - 44
14 years ago

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