The slope of the tangent to f(x) is
f'(x) = (g'h-gh')/h^2
So, the normal has slope h^2/(gh'-g'h)
At x=2, that is 6^2/(18*2-5*6) = 6
f(2) = g(2)/h(2) = 18/6 = 3
So, now you have a point and a slope, so the normal line at x=2 is
y-3 = 6(x-2)
let f(x)= (g(x)/h(x) where g(2)=18, h(2)=6, g'(2)=5 and h'(2)=2. Find the equation of the normal to the graph of f at x=2.
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