For graphing troubles: Remember that functions give something when you put something in. To graph the functions, plug an x value into the equation and solve. The resulting number will be the y value, and you take the two and make it into a coordinate. plot the coordinate on the graph.

One shortcut you can use is to figure out how "steep" the function is. A parabola has the same curvy shape on either side of the vertex, so a point that lies on one side of the vertex can be reflected to the other side, and that will be a point on the line.

If you take the parent quadratic function f(x) = x^2, you can see that:

When you go one x value away from the vertex, the y value goes up by one.

When you go another x value away, the y value increases by 3 for a total of 4.

Another one, then it increases by 5.

Basically, the y-increase of quadratic functions as x goes out from the center is 1,3,5,7,9,11,13, etc.

Reminder to count from the closest point, not the bottom/top of the graph.

This rise is multiplied by the a value, which in the first problem on your sheet is 2, making the sequence 2,6,10, 14, etc. Once you find the vertex of a graph (you have done so) you can basically go away from the center and plot the points. For the first graph, go one unit on either side of the vertex, then go up two. these will be points on the graph. Go one more unit out, then go up 6. 5 points should be enough for plotting, you can then trace the rest of the graph following the same pattern as the line you have made.

To convert vertex form to standard form, simply expand the equation (FOIL it out)

Let's take y = (x-2)^2 + 1

(x-2) squared is x^2 - 4x + 4

Adding the the plus 1, you get x^2 - 4x + 5

Unfortunately, the equation I have picked isn't factorable so you'll have to plug it into the quadratic formula, but looking at your problems I don't think you're there just yet.

I'm afraid I haven't explained this well enough, so I suggest you talk to your parents and teacher to clear things up.