There was an issue where I was placed in the wrong class at first this year and it's only just been rectified. I'm now in my proper Math 30 Pure class. My teacher was kind enough to waive the first few cumulative reviews and tests, but I was just assigned one that is going to be weighted the same as all of those past ones combined. Fun. The problem is that I've literally been to 3 classes, so while I've got a grasp of 2 lessons (one day was assigned to work on this review), I'm at a loss for the other 20 or so that the class has covered in my absence. Here are a few problems that I'm having issues with.

Given the function y= x^2(x+3)-6, I need to make the following transformations.

-Stretch it horizontally by 1/3

-Reflect about the x axis

-Reflect about the y axis

-Move it up 6 units, left 4.

From just messing around with my TI-83, as well as prior knowledge, I THINK that this is the proper solution, but feel free to correct me if I'm wrong.

- y= 3x^2(3x+3)-6

- y= -1(3x^2(3x+3)-6)

- y= -1(-3x^2(-3x+3)-6)

- y= -1(-3x^2(-3x+7))

Like I said, feel free to eviscerate me if I'm wrong on any of these.

Next, there's this.

Given 1/2y= f(-3x-9)+2, I need to find out what the points

(x,y)

(-6,-3)

(0,7) would change to if the equation was changed to

y=(f(x))^-1

I believe that this is an inverse function (like I said, could be wrong. It might be a reciprocal), which means you switch x and y, right? So

(y,x)

(-3,-6)

(7,0)

Like I said though, not sure if it's inverse or reciprocal.

Then there's this.

Given f(x)=√(x)+7, determine

x=f(y)

X intercepts for y+f^-1(x)

Invariant points for y=f(x) and y=f^-1(x)

On this one I'm just lost.

Finally, I need to find the formula for coterminal angles for

25π/8 and

8.3

Both of those in radians, of course. I understand the coterminal formula is and angle + n360, nEI, but that's for degrees. How do I do it for radians?

Again, any help for any of this would be greatly appreciated. Math has never been my strongest suit, but it's even harder when I haven't had a chance to learn it.