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 $2$

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