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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Yendall on March 14, 2013, 07:37:29 pm

Title: Finding points of a graph (on behalf of a friend)
Post by: Yendall on March 14, 2013, 07:37:29 pm: My friend was wondering how to do this question:

The graph of $y=2Sin^-^1(2x-1)+1$ is shown below.
Determine the values of a, b and c

(http://i.imgur.com/Uy6Ey0o.jpg?1)
Sorry about the shitty quality graph.

My friend and I would appreciate the help! thaaaanks
Title: Re: Finding points of a graph (on behalf of a friend)
Post by: Daenerys Targaryen on March 14, 2013, 07:44:09 pm: For b you can say. when x=0 y=b
then solve for b.

for the a and c. I dont see much of what you can do. unless that point is the turning point then you could do it. But i dont wanna assume anything
Title: Re: Finding points of a graph (on behalf of a friend)
Post by: Lasercookie on March 14, 2013, 07:58:51 pm: Quote from: Yendall on March 14, 2013, 07:37:29 pm
The graph of $y=2Sin^-^1(2x-1)+1$ is shown below.
So we'll basically want to find the domain and range of the function we're given. The domain will be of the form $[0,a]$ and the range will be $[b,c]$ , which allows us to find out answers.
We can look at the inverse sine function without any transformations, and then look at what transformations have be done to the function we've given.

So as Haters pointed out, we can find b quite easily.

$b = y = 2arcsin(-1)+1 = 2\frac{-\pi}{2} + 1 = 1 - \pi$

We know the domain and range of inverse sine (yay spesh formula sheet):
Domain of arcsin(x): $[-1,1]$

Range of arcsin(x): $[-\frac{\pi}{2},\frac{\pi}{2}]$

So domain of y: $[0, 2x-1] = [0, 2(1)-1] = [0,1]$

which means a = 1

Range of y: Well instead of looking at this from the point of view of transformations, we can sub in our domain values now, which we can see will be an okay thing to do looking at the graph and noting that it's increasing.
$[b,c] = [2arcsin(-1)+1, 2arcsin(2-1)+1] = [1 - \pi,2arcsin(1)+1] = [1 - \pi,2\frac{\pi}{2}+1] = [1 - \pi,1 + \pi]$

So $b = 1 - \pi$ and $c = 1 + \pi$