Method
Using the quadratic lines y = x , y = 2x , y = 3x , y = 4x , and y = 5x , I will plot these onto graph paper to get 5 curves. I will then work out the gradient at the points, x = 1, x = 2, x = 3,
x = 4, and x = 5 on those graphs, using two integer points on a tangent parallel to the curve at the above points and the formula, ‘y co-ordinate of B – y co-ordinate of A / x co-ordinate of B – x co-ordinate of A’. I will then tabulate all of those gradients, and from that table hopefully work out the common rule.
Graphs
On separate sheets.
Results
Curve x = 1 x = 2 x = 3 x = 4 x = 5
y = x 2 3.3 6 8 10
y = 2x 3.6 6.6 12.5 13.3 28
y = 3x 5.5 11.1 20 21.6 50
y = 4x 8.3 14.3 22 30 55
y = 5x 10 18.8 24.4 38 64
I then compared my results with other peoples’ results and made this table.
Curve x = 1 x = 2 x = 3 x = 4 x = 5
y = x 2 4 6 8 10
y = 2x 4 8 12 16 20
y = 3x 6 12 18 24 30
y = 4x 8 16 24 32 40
y = 5x 10 20 30 40 50
I will be using this table to find out the rule
Analysis of my results
The number after x multiplied by the number in front of x gives you the gradient for that curve at point x = 1. If we call the number after x ‘r’, the gradient ‘Gr’, and the number before x is ‘a’ therefore
‘ra = Gr at x = 1’ Let’s test this.
y = x . r = 2 a = 1 2 x 1 = 2 ‘Gr at x = 1’ = 2
y = 2x . r = 2 a = 2 2 x 2 = 4 ‘Gr at x = 1’ = 4
This equation works.
How do I get to x = 2?
ra + ‘Gr at x = 1’ = Gr at x = 2 Let’s test this.
y = x . r = 2 a = 1 Gr at x = 1 = 2
(2 x 1) + 2 = 4 = Gr at x = 2
y = 2x . r = 2 a = 2 Gr at x = 1 = 4
(2 x 2) + 4 = 8 = Gr at x = 2
This equation works for finding x = 2.
How do I find x = 3/4/5?
If I multiply ra by x, will I come to the gradient for x = 3?
y = x . r = 2 a = 1 x = 3
2 x 1 = 2 2 x 3 = 6 6 = Gr for x = 3
I think I’ve found the rule!!
rax = Gr
Let’s test it with y = 4x .
y = 4x . r = 2 a = 4 x = 2
2 x 4 = 8 8 x 2 = 16 16 = Gr at x = 2
y = 4x . r = 2 a = 4 x = 3
2 x 4 = 8 8 x 3 = 24 24 = Gr at x = 3
THIS RULE WORKS!!!
Conclusion
To work out the gradient at any point on my five graphs the equation ‘rax = Gr’ works perfectly.
Evaluation
I found that my results were slightly off from what they should be, but I compared my results with the results of others and corrected that error. The assignment had a very small margin for error, but a HUGE risk of it happening because the tangents might not be perfectly exact to the curve, as it relies upon the human eye to determine where the curve is and what direction the tangent should point in, and also the hand to draw the curve in the first place.
