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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#

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.

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Answer 2

Samantha Adkison

To find the slope of the secant line of (y = 2x^2 + \sec(x)) at (x = 0) and (x = \pi), first, find the corresponding y-values by substituting (x = 0) and (x = \pi) into the equation. Then, use the formula for slope:

[ \text{Slope} = \frac{\text{change in } y}{\text{change in } x} ]

Subtract the y-values and x-values obtained for (x = 0) and (x = \pi) and use them in the slope formula to calculate the slope of the secant line.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.