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

1 Answer
Apr 2, 2017

#(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.