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First, I went through each of the statements and picked which graphs agree with them. For “a)”, I realised that the turning point must have an x value which equals to 3. The two graphs which had this were 1 and 9. The next stated that the graph it corresponds to has a non-integer root, which only corresponds to 6, which has the roots, -1 and 1.5. C states that its graph has a line of symmetry which is x=k, where k<0. This shows that the vertex must be on the left side on the y-axis, and the only graph which has this is 5. D says that all y values must be above 0 and the only graph with this is 8. E says that the vertex of its graph line on x=1, but multiple graphs fit this statement. 2, 3, 7 and 8 all have a turning point on the line x=1 but since we have already used 8, we can rule it out. This leaves 2,3 and 7. Next comes F, which shows that its graph has a constant term of -8, so this shows that its roots must make a product of -8. The three graphs which have this are 3 and 7 (which have the roots -2 and 4) and 5 (which have the roots -4 and 2). We can rule out 5 as we have already used it and so are left with 7 and 3. G states it has a graph which the sum of its roots is 6. The only one which makes this statement true is 1 (which has roots 1 and 5). H says the two points (0,8) and (2,8) lie on its curve and the only graph with that is 7. Lastly, we have I which says the sum of its roots is an odd number (b is odd). As there are only two roots in a quadratic graph, we must find the graph with an even and odd root. This will make a sum of an odd number and the only graph that fits this requirement is 4.
After going through all statements, we must go back to the ones we haven’t solved. These would be A, E and F. Going back to A we see we have either 1 or 9, but as 1 has already been used, we get the answer 9. Next, we have E which we got the answers 7, 3 and 2. 7 is already used so we are left with 3 and 2 but cannot make a decision between the two. We will have to go onto F, which answers were 7and 3. Again we can rule out 7, as it has already been used, which leaves us with 3 as the answer. Going back to E, we can rule out 3 which leaves us with 2.
This leaves us with the answers A-9, B-6, C-5, D-8, E-2, F-3, G-1, H-7 and I-4.