How do you find the slope of the secant lines of #f(x)=6x-1x^2# through the points: #(2, f (2))# and #(2 + h, f (2 + h))#?

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Answer 1 · 2016-09-25T11:01:17

#2-h#

The line slope is given by

#m(h) = (f(2+h)-f(2))/(2+h-2) = (f(2+h)-f(2))/h#

f=6x.x^2or substituting values

#m(h) = ((6-(2+h))(2+h)-(6 xx 2-2^2))/h = 2-h#.

Now if you calculate

#lim_(h->0)m(h) = lim_(h->0) 2-h = 2#

which is the coeficient of the tangent line to #f(x)# at point #2#