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April 21st, 2014, 04:47 PM   #1
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Help with Functions?

Hi, first off, sorry if this is the wrong section, hopefully it's right.

Anywas, I'm having troubles with finding out what exactly functions are. I never took Pre-Calculus 11, and I'm currently taking Pre-Calculus 12. What exactly is a function? I've tried looking it up on google, but it doesn't really make sense to me. Every question that my teacher has assigned me has the ƒ on it, but I don't really understand what's the purpose of it? Some examples of questions I have (Not exactly the same as the ones I have, but close enough.) are and . What steps do I take to solve those?

I'm just really confused about functions and have no idea what they mean. Any help would be appreciated. Thanks. (Also, it's an online course I'm taking if that helps or anything, the only reason I'm taking Pre-Calculus 12 is because it's required for the program I wish to go into next year, and I can't find any online courses offering Applications of Math 12 which they also accept. Also, is it likely to finish Pre-Calculus 12 within 6~ weeks?)

Thanks once again.

Last edited by xKy; April 21st, 2014 at 04:49 PM.
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April 21st, 2014, 07:10 PM   #2
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A function simply maps one set of numbers (the domain) onto another (the range) in a defined way. The function $f$ defines an operation that is performed on a number from the domain, and the resulting number is in the range.

If $x$ is a number in the domain of the function, then $f(x)$ is the resulting number in the range. To represent the function on a graph, we draw a curve on which every point $(x,y)$ has the property that $y = f(x)$. This, then, is the graph of $y = f(x)$.

In your example, you are shown one graph $y = f(x)$ and are asked to draw $y = f(\frac{1}{2}x)$. To do this, we look at the points on the original graph and look at how we need to change the coordinates.

So, from the left: there is a point at $(-3,0)$. This represents the point $y = f(-3) = 0$. In our new graph, we need to put $\frac{1}{2}x$ into the function instead of $x$ to get the same value of $y$. Thus $$\frac{1}{2}x = -3 \; \Longrightarrow \; x = -6$$
And so we mark the point $(-6,0)$ on the graph.

We do the same with the next point at $(0, -2)$. This time we get $$\frac{1}{2}x = 0 \; \Longrightarrow \; x = 0$$
And so we mark $(0, -2)$ on our graph. We can join the two points using a curve that follows the shape of that on the original graph of $y = f(x)$.

There are two more points to mark (I'll let you work them out), but the lines connecting them are straight, so those on our graph will be straight too.
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April 21st, 2014, 10:59 PM   #3
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Thanks for the response, sounds really confusing as I read it over, but I'm beginning to understand a bit.

So, for the question I substitute Y for f(1/2x)? And would it be the same for every other question (Substituting the Y/X and the function)?

And I still don't understand how you got-6 on the left most point and the 0. How did you solve / get that number?
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