I got the first equation by using the graph. I plotted -1 1/2 on the x axis and -1/2 on the y axis and drew the line. I then labelled it y=x+1.

I can't work out how you got the second onethough.
Don't plot x=-1.5 and then plot y=-0.5. You're supposed to find the point (-1.5,-0.5) on the graph. When you find that point (on the graph provided), you will see that two of the lines on that graph intersect that point. The two lines that are intersecting at that point are the answers for the problem.

I got the first equation by using the graph. I plotted -1 1/2 on the x axis and -1/2 on the y axis and drew the line. I then labelled it y=x+1.
that is incorrect. not only is your method off (as you don't have to draw anything for these questions) but the line you drew would not be y = x + 1

ecMathGeek gave a short explanation for this question, look at it

I can't work out how you got the second onethough.
They gave the solutions as -3.3 or 0.3 so i looked at the graph for those x-values to see if any graphs intersected at that those points. i found that the quadratic and the line y = -x - 2 intersected there, so i just equated them (since intersecting points occur when one graph equals the other) and simplified the answer to look like a quadratic. when i used the quadratic formula, i was just verifying my answer, you would not have to do that.......do you understand now, or is something still unclear?

3. OK. Here goes:

(a) y=x+1 and y=-x-2
Got it. Thanks.

(b) x = -3.3 or 0.3
Looked at the graph, found that the following two lines interestected at those points. So they equal each other at those points:

y= x^2 + 2x -3 and y= -x -2

x^2 + 2x -3 = -x -2

x^2 + 3x -1 = 0

Which is the answer given by you guys.

The answers with the paper give a mark for this, but the final answers is given as:

x^2 + 2x - 3 = x + 1

-How did they get that?

(c) I read what you wrote, but still cant see how to give the solutions from the graph.

OK. Here goes:

(a) y=x+1 and y=-x-2
Got it. Thanks.
good

(b) x = -3.3 or 0.3
Looked at the graph, found that the following two lines interestected at those points. So they equal each other at those points:

y= x^2 + 2x -3 and y= -x -2

x^2 + 2x -3 = -x -2

x^2 + 3x -1 = 0

Which is the answer given by you guys.

The answers with the paper give a mark for this, but the final answers is given as:

x^2 + 2x - 3 = x + 1

-How did they get that?
i think you are looking at the wrong question, that's what we used for part (c). if you check the solutions to that equation, you will realize it is not -3.3 or 0.3

(c) I read what you wrote, but still cant see how to give the solutions from the graph.
we realize that we can construct the given quadratic using two of the original graphs given

we can take y = x^2 + 2x - 3 and y = x + 1 and subtract them, the result will be the given quadratic.

x^2 + 2x - 3 - (x + 1) = x^2 + x - 4

now we want to solve x^2 + x - 4 = 0

that means we want x^2 + 2x - 3 - (x + 1) = 0

but that's the same as x^2 + 2x - 3 = x + 1

and so the answers will be where the graphs y = x^2 + 2x - 3 and y = x + 1 intersect

5. The final answer for (b) in the exam paper is definitely what I posted. Maybe it's a misprint and should have been with the next question.

Anyway, You said that we can construct the given quadratic using two of the lines from the graph. How would you know which ones to use and whether to subtract or add? Trial and error?