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I think this maybe a basic question, but I'm not mathematically equipped to handle the details.

The problem I have is that I will have a curve such as in the image below and I want to find the y-axis value in relation to the curve from a known value along the x-axis.

I'm also somewhat unsure how to plot a curve of that type below, any pointers would be a bonus for me.

enter image description here

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Given the nature of the question, I'm dubious that "bezier-curve" is an appropriate tag. See if you are interested in a (very rough) intro to Bezier Curves. – Shaun Ault Jan 5 '13 at 14:35
I couldn't tag it with curve, I don't know what sort of curve it is... – JDandChips Jan 5 '13 at 20:44

1 Answer 1

Let's assume that you have an equation for the curve, let's say for example, $y = \sqrt{x}$. The this equation is what defines the relationship between $x$-values and $y$-values. Given a known $x$-value, you just have to plug that value in for $x$ in the equation to find the $y$-value. In the example $y = \sqrt{x}$, When $x = 100$, the $y$-value is $y = \sqrt{100} = 10$.

If you are trying to plot an unfamiliar function, you may want to organize your work into a table of sample $x$-values and $y$-values. List a few $x$-values, and then use your equation to determine the appropriate $y$-values. Here's an example for the equation $y = \sqrt{x}$.

First, choose some $x$-values: $$ \begin{array}{l|l} x & y = \sqrt{x} \\ \hline -1 & \\ 0 & \\ 1 & \\ 2 & \\ 3 & \\ 4 & \\ \end{array} $$

Then, find the $y$-values, based on the given equation: $$ \begin{array}{l|l} x & y = \sqrt{x} \\ \hline -1 & \sqrt{-1} = \textrm{not a real number}\\ 0 & \sqrt{0} = 0\\ 1 & \sqrt{1} = 1\\ 2 & \sqrt{2} \approx 1.414\\ 3 & \sqrt{3} \approx 1.732\\ 4 & \sqrt{4} \approx 2\\ \end{array} $$ This tells you that some points on the graph include $(0,0)$, $(1,1)$, $(2,1.414)$, $(3,1.732)$, $(2,2)$. In fact, there are infinitely-many points on the curve, be we can often get a good idea of the shape of a simple graph based on a few points.

Hope this helps!

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