Question

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

Answer 1

`(2pi^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*0^2+sec(0) = 0 + 1/(cos(0)) = 1`,
and
`x_2 = pi`
`y_1 = 2*pi^2+sec(pi) = 2pi^2 + 1/(cos(pi)) = 2pi^2-1`

The slope `k` of a straight non-vertical line is given by the formula
`k = (y_2-y_1)/(x_2-x_1) = (2pi^2 - 1 - 1)/(pi -0) = (2pi^2 - 2)/pi`,
which in this case is the slope of the sought secant line.