Hey! Sorry for my late response!
For Q1, can you please specify whether the question is asking for the inverse of (1/x)+1 or 1/(x+1)? It's not exactly clear when typed haha!
But whatever the original equation is, remember when finding inverse you can
equate any of the equations to x, because the inverse and the original graph ALWAYS intersects along the line y=x. Don't bother trying to equate them to each other.
As for implied domain of (x-1)/(x+2), IDK how others would do it but I would first convert it to -3/(x+2)+1, so from this you can tell that the asymptotes will be at y=1 and x=-2. Remember that the graph never touches the asymptotes, therefore you can deduce that the implied domain would be R\{-2}.
As for sketching, I usually start with the asymptotes, so draw two dotted lines for y=1 and x=-2. Then look at what shape the graph would be and whether or not it would have any x- or y-intercepts. Then draw the lines.
For graphing - it's good to know what shape each type of graph would be. Look at the power to know what type of graph it is. For example, in Q2 it's 16-x^2. A power of two implies it's a
quadratic, which is a parabola. Then look at what form the graph is written in. For 16-x^2, you can rewrite this as (4-x)(4+x), so you can find out its x-intercepts. If it was written in ax^2 + bx + c, you can find the y-intercept. Finally, if it was a(x+b)^2 + c, you can find the turning point.
What I use to remember is that the power implies how many x-intercepts a graph could have up to. A linear (x^1) only has one intercept, quadratic (x^2) has up to two x-intercepts. A cubic (x^3) has up to three, and a quartic (x^4) would have up to four etc.
Hopefully that clears it up for you
Lemme know if you have any more questions!