Calculating Y = Sec(X) Functions Given Points?

Calculating Y = Sec(X) Functions Given Points?
$\begingroup$

I'm attempting to write a function for a computer program that will allow me to find points along it. To start with, I have 2 points. One is always on the y axis, and the other is always larger in both coordinates.

I started with an exponential function of the form $y = e^{kx}$ where I calculated points with $$k = \frac{ln(|y_1|)-ln(|y_0|)}{x_1-x_0}$$

This worked fine for a period but then I wanted to change how aggressive the curve was.

I realized I could calculate a linear function with these points as well, and combining the linear and exponential functions works nicely for creating a function through the same points.

$$y = a(Ce^{kx})+(1-a)(mx)$$ $$0\le a\le1$$ Now in order to make the curve less aggressive at the beginning, but still pass through both points, I think I might be able to use a $y = sec(x)$ function.

The problem is that I can't seem to figure out how to calculate a $sec(x)$ function given the points that I have.

Here is a desmos graph showing some of the thinking I've done about this problem, adjusting the ($a$) parameter can show you how the function(s) can become more or less aggressive. Right now, the $sec(x)$ function isn't calculated and is only eyeballed for graphing.

Is there a way to calculate the multiple of x in the $sec(x)$ function based off the points I have? Or is there another function other than a $sec(x)$ function that could work similarly?

Thanks for your time.

$\endgroup$
1

1 Answer

$\begingroup$

With some help, I managed to figure this out. Given a $y$ intercept $(0, y_0)$ and another point $(x_1, y_1)$ one can calculate Linear, Exponential, and Secant functions as follows:

Linear

$$y=(\frac{y_1-y_0}{x_1})x+y_0$$

Exponential

$$k=\frac{ln(abs(y_1))-ln(abs(y_0))}{x_1}$$ $$y=y_0e^{kx}$$

Secant

$$s = \frac{arccos(\frac{1}{(y_1-y_0)+1})}{x_1}$$ $$y = sec(sx)+y_0-1$$

Try these interactively on my updated desmos:

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Chloe Bennett
Author

Chloe Bennett

Chloe Bennett explores the intersection of pop culture, streaming entertainment, digital trends, and contemporary lifestyle. Her weekly commentary reaches thousands of culture enthusiasts.