my sister is really stuck on these Algebra questions, she needs major help she goes to UCSB, MAuhahA I'm evil

Question

Write the function whose graph is the graph of y = x3, but is shifted to the left 2 units.
Y= ???
Write the function whose graph is the graph of y = x3, but is shifted up 9 units.
Y=???
Write the function whose graph is the graph of y = x3, but is shifted down 6 units.
Y=???
Write the function whose graph is the graph of y = x3, but is reflected about the y-axis.
y=???
Write the function whose graph is the graph of y = x3, but is reflected about the x-axis.
y=???
Write the function whose graph is the graph of y = x3, but is vertically stretched by a factor of 3.
y=???
Find the function that is finally graphed after the following transformations are applied to the graph of . Note: In Step 2 both the transformation in Step 1 and the transformation in Step 2 take place. In Step 3 all three transformations take place.
1) Shift up 8 units Y=???
2)Reflect aboukt the x-axis Y=???
3)Reflect about the y-axis Y=???
Help her this is only the few out of 50 questions she stuck on.

Answer 1

When the graph is shifted to the left, this means that the value the function had at x = 0 now occurs at x = -2. To do this, change x to x+2. Then when x = -2, the new function will have the value of the old ƒ(0).

When shifted up, the value of y is three units more than previous, or y(new) = y(old)+3, therefore shifted up y = y + 3 = x^3+3

You should see the third part now; it's just a different sign.

For a function to be reflected about the y axis, ƒ(-x) = ƒ(x). Therefore we want x^3 to equal (-x)^3 for x<0. Since (-x)^3 is always negative, to make it the same as x^3, use -x^3 (x < 0).

Reflection about the x axis is just a sign reversal in y, so just take the negative of the function over its full range.

Vertically stretched means each value of y is mutiplied by a factor. In your problem, that factor is 3, so the stretched function is 3y or 3*x^3.

Finally, just apply the transformations one at a time. First, shift y up by adding 8 to the function, then take this new function and reverse its sign for x-axis reflection, then determine if the function ƒ(-x) = ƒ(x) for y-axis reflection and make the necessary adjustments.