... then, at seven, school starts. It's strange, I suppose, because I always look forward to the first lesson of the year. Especially this year because we have a new teacher called Mr. Pennyfeather. We haven't seen him yet, none of ...

... task is simply asking for where the tangent line intersects the cubic function when the tangent point is at "x=-2.25". Before finding the function for the tangent line, we will need to first find its "y" value, simply by substituting ...

... give (197, 16.5) When t = 197, the model gives y = 6.6 sin (0.986(197 - 113)) + 10.2 y = 6.6 sin (82.824)+10.2 = 6.6 x 0.992 + 10.2 = 16.75 is very close to the data value of 16.7, ...

... - y1 = m(x - x1) (where m is the gradient) to obtain the required equation. Co-ordinates of the mid-point of OA are ( ) = (1,3) The gradient of the line joining the points O (0,0) and A(2,6) is a measure of ...

... method is not very accurate though and large discrepancies in gradient can occur. This is because this method involves manually drawing a tangent which will often be drawn incorrectly. Errors in the measurement of changes in Y and X can ...

... and 'n' are constants, and then investigate the Gradient function for any curves of my choice. I will first find the gradient of tangents on the graph y=x² by drawing the graph (page 3), and ...

... 1sf) At (2.8) 15 1.3 11.538462 = 12(to 1sf) At (3,27) 20 0.8 25 At (4,64) 22 0.5 44 I haven't noticed a pattern in this set of results, but I think that at (3,27) and (4,64), my gradients are a bit wrong. I think it is that the second difference goes up ...

... x gradient 1 1 3 1 3 1.1 1.21 2.79 0.9 3.1 1.2 1.44 2.56 0.8 3.2 1.3 1.69 2.31 0.7 3.3 1.4 1.96 2.04 0.6 3.4 1.5 2.25 1.75 0.5 3.5 1.6 2.56 1.44 0.4 3.6 1.7 2.89 1.11 0.3 3.7 1.8 3.24 0.76 0.2 3.8 1.9 3.61 0.39 0.1 3.9 1.99 3.9601 0.0399 0.01 3.99 1.999 3.996001 0.003999 0.001 3.999 2 4 2.001 4.004001 -0.004001 -0.001 4.001 2.01 4.0401 -0.0401 -0.01 4.01 2.1 4.41 -0.41 -0.1 4.1 2.2 4.84 -0.84 -0.2 4.2 2.3 5.29 -1.29 -0.3 4.3 2.4 5.76 -1.76 -0.4 4.4 2.5 6.25 -2.25 -0.5 4.5 2.6 6.76 -2.76 -0.6 4.6 2.7 7.29 -3.29 -0.7 4.7 2.8 7.84 -3.84 -0.8 4.8 2.9 8.41 -4.41 -0.9 4.9 3 9 -5 -1 5 Power: 2 Coefficient: 1 Fixed point: 2 My second fixed point is 5, 25 x y change in y change in x gradient 4 16 9 1 9 4.1 16.81 8.19 0.9 9.1 4.2 17.64 7.36 0.8 9.2 4.3 18.49 6.51 0.7 9.3 4.4 19.36 5.64 0.6 9.4 4.5 20.25 4.75 0.5 9.5 4.6 21.16 3.84 0.4 9.6 4.7 22.09 2.91 0.3 9.7 4.8 23.04 1.96 0.2 9.8 4.9 24.01 0.99 0.1 9.9 4.99 24.9001 0.0999 0.01 9.99 4.999 24.99 0.009999 0.001 9.999 5 25 5.001 25.01 -0.010001 -0.001 10.001 5.01 25.1001 -0.1001 -0.01 10.01 5.1 26.01 -1.01 -0.1 10.1 5.2 27.04 -2.04 -0.2 10.2 5.3 28.09 -3.09 -0.3 10.3 5.4 29.16 -4.16 -0.4 10.4 5.5 30.25 -5.25 -0.5 10.5 5.6 31.36 -6.36 -0.6 10.6 5.7 32.49 -7.49 -0.7 10.7 5.8 33.64 -8.64 -0.8 10.8 5.9 34.81 -9.81 -0.9 10.9 6 36 -11 -1 11 Power: 2 Coefficient: 1 Fixed Point: 5 My third fixed point is -3, 9 x y change in y change in x gradient -4 16 -7 1 -7 -3.9 15.21 -6.21 0.9 -6.9 -3.8 14.44 -5.44 0.8 -6.8 -3.7 13.69 -4.69 0.7 -6.7 -3.6 12.96 -3.96 0.6 -6.6 -3.5 12.25 -3.25 0.5 -6.5 -3.4 11.56 -2.56 0.4 -6.4 -3.3 10.89 -1.89 0.3 -6.3 -3.2 10.24 -1.24 0.2 -6.2 -3.1 9.61 -0.61 0.1 -6.1 -3.01 9.0601 -0.0601 0.01 -6.01 -3.001 9.006001 -0.006001 0.001 -6.001 -3 9 -2.999 8.994001 0.005999 -0.001 -5.999 -2.99 8.9401 0.0599 -0.01 -5.99 -2.9 8.41 0.59 -0.1 -5.9 -2.8 7.84 1.16 -0.2 -5.8 -2.7 7.29 1.71 -0.3 -5.7 -2.6 6.76 2.24 -0.4 -5.6 -2.5 6.25 2.75 -0.5 -5.5 -2.4 5.76 3.24 -0.6 -5.4 -2.3 5.29 3.71 -0.7 -5.3 -2.2 4.84 4.16 -0.8 -5.2 -2.1 4.41 4.59 -0.9 -5.1 -2 4 5 -1 -5 Power: 2 Coefficient: 1 Fixed point: -3 Observation For the equation Y=X2, the ...

... will then do other functions such as Y=4x2 Next I will move to a function y = x4 and investigate the gradient at different points. I will use the same method used in the equation y = x2 but instead of squaring ...

... intersect the curve, as this will distort the outcome. As a result, we can say that the curve's steepness at a particular point is identical to the tangent formed in that specific curve. I will draw four graphs: y=x2, y=x3 ...

... it is more accurate than the tangent one. You zoom into the graph and take the x point and y point just after a whole figure. For instance with y-x2 , using point (1,1) we would do 1.0012 - 1 ...

... y=x2 See Graph 1 x -3 -2 -1 0 1 2 3 gradient of curve -6 -4 -2 0 2 4.27 6.36 Observation: I noticed that the gradient of the curve is approximately 2 times the value of x. Then I looked at the curve y=ax2. See graph 2 x -3 -2 -1 0 1 2 3 Gradient of curve -6a -4a -2a 0 2a 4a 6a Observations: Everything is times a because it is ...

... x value. Therefore, for the graph of y=x2, the formula for the gradient of a tangent is g=2x. I am left with the question of accuracy. I cannot get total accuracy, but there are ways I can get very close to ...

... using a formula that is provided in the 'Essential Pure Mathematics' A-level textbook. The formula is:- RQ = NQ - NR = (x + h)² - x² = 2xh + h² and PR = h The gradient of the chord PQ is:- RQ = 2xh ...

... Curves involving x3 + x 1. y = x3 2. y = 2x3 3. y = 4x3 + 2x - 5 Finally, I will summarise my results in a series of tables and work out an overall formula that I could use to predict the gradient ...

... graphs and the gradient function at each point in a curve is different therefore I will draw tangents using a capillary tube and calculate the gradient by taking the change in the value of the y co ordinates divided by ...

... of J= increase in y-axis/increase in x-axis = 3/1 =3 Gradient of K = increase in y-axis/increase in x-axis = 3/1 =3 Gradient of L = increase in y-axis/increase in x-axis = 3/1 =3 5. y=-3x graph Gradient of M = increase in y-axis/increase ...

... 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 ...

... at the graph of y=x2. See graph B By drawing the tangents of each point, we can calculate the gradients. However, as the graph is not always completely accurately drawn, there is likely to be some error in the results. x g by ...

... Graph --> y=3x3 Small increase method Fixed point A= (2,24) B1= (2.01,24.361803) B2= (2.001,24.036018) B3= (2.0001,24.00360018) AB1 Gradient = 24.361803-24 2.01-2 = 0.361803 0.01 = 36.1803 AB2 Gradient = 24.036018-24 2.001-2 = 0.036018 0.001 = 36.018 AB3 Gradient = 24.00360018-24 2.0001-2 = 0.00360018 0.0001 = 36.0018 The Gradient is approaching 36 Graph --> y=3x3 Small increase method Fixed point A= (3,81) B1= (3.01,81.812703) B2= (3.001,81.081027) B3= (2.0001,81.00810027) AB1 Gradient = 81.812703-81 3.01-3 = 0.812703 0.01 = 81.2703 AB2 Gradient = 81.081027-81 3.001-3 = 0.081027 0.001 = 81.027003 AB3 Gradient = 81.00810027-81 3.0001-3 = 0.00810027 0.0001 = 81.00270003 The gradient is approaching 81 Tabulating The Results For Graph ...

... as they are in maths. In physics acceleration is the gradient of a velocity time graph and velocity is the gradient of an instance time graph. Also gradients can be used in radioactivity and decay in physics. In biology and chemistry ...

... be a2 against Cos?. On this graph, three lines will be drawn: a line of maximum gradient, a best fit line and a line with minimum gradient going through the points. The gradient (m) will equal the part of the ...

... X3). So if the formula for the gradient is Height/Width then, by replacing the height with AXN and the width with X/N we get XN/(X/N). We can simplify this by multiplying both sides by N to get ANXN/X and we ...

... you can see, they are all twice their x value. Therefore, for the graph of y=x2, the formula for the gradient of a tangent is g=2x. I am left with the question of accuracy. I cannot get total accuracy, ...

... the graph 'y= x3' is 3x2. However, I can also conclude that the method of using tangents to measure the gradient is too inaccurate to produce good results. Therefore, in my next investigation I will use the small increment method. ...