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

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

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Answer 1 · 2017-04-02T11:29:32

$\frac{2 π^{2} - 2}{π}$ $(2pi^2 -2)/pi$

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:
$x_{1} = 0$ $x_1 = 0$
$y_{1} = 2 \cdot 0^{2} + sec (0) = 0 + \frac{1}{cos (0)} = 1$ $y_1 = 2*0^2+sec(0) = 0 + 1/(cos(0)) = 1$ ,
and
$x_{2} = π$ $x_2 = pi$
$y_{1} = 2 \cdot π^{2} + sec (π) = 2 π^{2} + \frac{1}{cos (π)} = 2 π^{2} - 1$ $y_1 = 2*pi^2+sec(pi) = 2pi^2 + 1/(cos(pi)) = 2pi^2-1$

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

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

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Explanation:

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How do you find the slope of the secant lines of y=2x2+sec(x)y=2x^2+sec(x) at x=0x=0 and x=π?x= pi?

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