Question 4 and 5. i don’t understand the simplifying for the fraction and i don’t understand graphing function.

(4a)

to find  is to find the inverse of 

steps to find inverse:
replace function notation with y
swap all x with y
solve for y
replace the y with 



replace function notation with y

   

swap all x with y (this is what actually makes it the inverse)

   

solve for y
multiply both sides by y+3 so denominator cancels on right-hand side
then distribute x into (y+3) on the left-hand side

   

move all y-terms together.
subtract both sides by 1

   

subtract both sides by xy
   
   

factor out y from the right side

   

divide both sides by (2-x)

   

replace y with 

   



(4b)

the function on an inverse function cancels each other out.
they "undo" one another and gets you back the original input

   

(4c)

   



(5a)

see attachment. is the yellow graph; it is a parabola with vertex (0,0) opening upward.
the dashed line is the line y=x.
the blue graph is the graph of  reflected in the line y = x.

when we reflect a graph of a relation  in the line y = x, that gets us the inverse relation of that original relation.
so there we have the inverse of y = x^2 in blue

(5b)

the function  does not have an inverse function.

from 5a that the (blue) inverse of the parabola  fails the vertical line test. it cannot be called an "inverse function" because its reflection, the inverse relation, not a function (it fails vertical line test)

(5c)

yes,  does indeed have an inverse function.

by restricting the domain like that from the vertex, we  basically chop the parabola in half.

when we reflect that it in the y = x line to get the inverse, that reflection will pass the vertical line test and it can be called an inverse function.

4, -4 is the other point it would pass through