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!