c Substitute the number of points into the equation found in Part B.
A
a
Number of Points, x
2
3
4
5
Number of Segments, y
1
3
6
10
B
b y=1/2x^2 - 1/2x
C
c 45
Practice makes perfect
a We are given a table that shows the number of segments y that we can draw between x non-collinear points.
Number of Points, x
2
3
Number of Segments, y
1
3
Let's draw a graph to illustrate the information given in the table.
As we can see, when there are two points we can draw only one segment, and when there are three points we can draw three segments. Let's make now a diagram with four points.
We can draw six segments when there are four points. We first draw a quadrilateral, and then draw its two diagonals.
Number of Points, x
2
3
4
Number of Segments, y
1
3
6
Next, let's make a diagram with 5 points.
As we can see, we can draw 10 segments when there are five non-collinear points. We draw a pentagon, and then a star connecting the rest of the vertices. We are ready to complete the table.
y = ax^2 + bx + c
Let's substitute three of the points on the table found in Part A into the equation above.
y=ax^2+bx+c
x
y
Substitute
Simplify
2
1
1=a( 2)^2+b( 2)+c
4a+2b+c=1
3
3
3=a( 3)^2+b( 3)+c
9a+3b+c=3
4
6
6=a( 4)^2+b( 4)+c
16a+4b+c=6
With the three obtained equations, we can form a system of equations.
4a + 2b + c = 1 & (I) 9a + 3b + c = 3 & (II) 16a + 4b + c = 6 & (III)
We can solve this system by using the Elimination Method. To do so, we will subtract Equation (I) from Equation (II) and Equation (III).
