Thread: Graphing 2008 HSC 3 a) iii)

1. Graphing 2008 HSC 3 a) iii)

I was attempting to graph the function from the 2008 HSC Q 3 A) iii) The way that I tried to tackle it was as follows:

1. Graphed y(-x), which is just a reflection of the graph about the y-axis.
2. Graphed y(-x + 2), which I thought should be the graph in step 1, slides to the left by 2 units, as the graph y(x-2) slides the graph to the right 2 units.

However, this was not the case.

Question: What was wrong in my thought processes? Could someone perhaps explain a better method of obtaining the graph?

Thanks

2. Re: Graphing 2008 HSC 3 a) iii)

y=g(x+2) is the graph of y=g(x) translated 2 units left, NOT y=g(-x+2)=g(2-x). In order to obtain g(2-x): Graph g(-x), which you were correct about this, and then translate this graph 2 units right, so it becomes y=g[-(x-2)]=g(2-x).
That's the theory behind it, but the shortcut to graphing f(a-x) when given f(x), the trick is just to reflect the graph about the line x=a/2 (that's the result of all the translations).
So what you did wrong was graph y=g(2-x) wrong, you need to translate g(-x) two units right, not left (or simply reflect y=g(x) about x=1). Following that simply draw that bit of the graph for x<1.

3. Re: Graphing 2008 HSC 3 a) iii)

Oh, ok. I think I understand now where what I thought was wrong, I was going define f(x) = g(-x), and then graph f(x+2) which would be sliding the graph f(x) to the left 2 units, but, that isn't g(-x+2) as f(x +2) = g(-(x+2)) = g(-x-2). I would have had to go f(x - 2) = g(-x +2) (as you said). Thanks

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