# 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))?

Sep 25, 2016

$2 - h$

#### Explanation:

The line slope is given by

$m \left(h\right) = \frac{f \left(2 + h\right) - f \left(2\right)}{2 + h - 2} = \frac{f \left(2 + h\right) - f \left(2\right)}{h}$

f=6x.x^2or substituting values

$m \left(h\right) = \frac{\left(6 - \left(2 + h\right)\right) \left(2 + h\right) - \left(6 \times 2 - {2}^{2}\right)}{h} = 2 - h$.

Now if you calculate

${\lim}_{h \to 0} m \left(h\right) = {\lim}_{h \to 0} 2 - h = 2$

which is the coeficient of the tangent line to $f \left(x\right)$ at point $2$