• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Investigating the Gradients of Graphs of the Form ‘y=ax’

Extracts from this document...

Introduction

Investigating the Gradients of Graphs of the Form 'y=ax' Aim To find a formula for the gradient of a 'y=ax' graph. To do this I will take a number of values of 'a' find the gradients and form a conclusion. Having done this I will test my theory with a different value of 'a'. The values of 'a' that I will test are: 1,2,3 and 5 Then a fraction and a negative number: 1/5 and -4. ...read more.

Middle

1 2 3 4 5 6 7 8 9 10 a=2 x 0 1 2 3 4 5 6 7 8 9 10 y 0 2 4 6 8 10 12 14 16 18 20 a=3 x 0 1 2 3 4 5 6 7 8 9 10 y 0 3 6 9 12 15 18 21 24 27 30 a=5 x 0 1 2 3 4 5 6 7 8 9 10 y 0 5 10 15 20 25 30 35 40 45 50 a=1/5 x 0 1 2 3 4 ...read more.

Conclusion

Conclusion I conclude that the gradient is equal to the value of 'a' and that therefore the gradient function equals 'a'. To test this conclusion I will draw a graph of 'y=8x' where 'a' equals 8. Results II a Gradient 8 8 Conclusion II This proves my theory so I can now say that in a graph of the form 'y=ax' the gradient equals 'a'. This can also be referred to as the gradient function(G.F.). G.F. = a ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Gradient Function section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Gradient Function essays

  1. Curves and Gradients Investigation

    y = x²) which can be used to precisely calculate the gradient at any point on the X - axis. This rule will be applied to the X value in question to give an exact gradient value for this point.

  2. Analysing Triangle Vertices and Bisectors

    The length of the perpendicular - labelled x on diagram - from point A to the line OB can be found by using the length of AB (10) and the angle ABO (26.6� (3s.f.)) - labelled ( on diagram. Because the perpendicular (x)

  1. The Gradient Function

    When it is only x in the curve, then the line will run through (1,1). When there is a 2x in the curve, the line will run through (1, 2). I am therefore going to make a prediction. When I demonstrate the curve of y=3x3, I am going to predict that the curve will run through (1, 3).

  2. I have been given the equation y = axn to investigate the gradient function ...

    co-ordinate is given ,e.g: At the point where x = 3, Gradient = 2x =2(3) =6 At the point (4,6) Gradient = 2x =2(4) = 8 Looking back at the longer method we used to find gradient at the point where x = 1 we see that the value obtained

  1. Investigate the gradients of the graphs Y=AXN

    This is good for I now know that my formula works for any cases of N. I shall now see if the A in the formula ANXN-1 is correct. I shall do this by keeping N as 2 as it is the lowest number and make A 2 then 3.

  2. The Gradient Function

    generalization easier I have drawn a table summarising the gradient found by the increment method at a common point in all the graphs. Equation X Gradient Y= 1/x -2 -0.25 Y=-2/x -2 0.5 Y= 3/x -2 -0.75 From the table above we can see that gradient function for the hyperbolas is '-a x (1/x)

  1. I am going to investigate the gradients of different curves and try to work ...

    finishes at the same point, its gradient will equal the gradient of a tangent drawn at the same point. From algebra I proved that for the graph y = 4x2 , the gradient of any chord is given by 8x+d.

  2. The Gradient Function

    This is extremely close to the answer that we get for differentiating sin 0.5 and therefore the derivative of sin x is cos x. However, I cannot show, apart from the algebra above, or explain why this works, although I have tried.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work