# How do you find the slope of the secant lines of y=2x^2+sec(x) at x=0 and x= pi?

Apr 2, 2017

$\frac{2 {\pi}^{2} - 2}{\pi}$

#### Explanation:

The secant line between two points on a curve (there is only one such line), is the straight line between the two points on the curve.

The two points here are:
${x}_{1} = 0$
${y}_{1} = 2 \cdot {0}^{2} + \sec \left(0\right) = 0 + \frac{1}{\cos \left(0\right)} = 1$,
and
${x}_{2} = \pi$
${y}_{1} = 2 \cdot {\pi}^{2} + \sec \left(\pi\right) = 2 {\pi}^{2} + \frac{1}{\cos \left(\pi\right)} = 2 {\pi}^{2} - 1$

The slope $k$ of a straight non-vertical line is given by the formula
$k = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{2 {\pi}^{2} - 1 - 1}{\pi - 0} = \frac{2 {\pi}^{2} - 2}{\pi}$,
which in this case is the slope of the sought secant line.