# The Gradient Function Maths Investigation

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Introduction

The Gradient Function

Introduction

A lot of graphs produce lines that are curves. Some of the curves are steep, and some are not.

In this investigation I will be looking to work out a formula, which will work out the gradient of any curve.

There are three methods for working out the gradient of a curve, all of them using a tangent:

- The Tangent Method
- The Increment method
- General Proof

This is the graph of y=x². I will find out the gradient of this curve, by using the three methods I mentioned above. I will use the point x=2 for this graph.

Tangent Method: The tangent method is already shown on the graph. I have drawn a line which touches the edge of the curve, when x=2. Once I have drawn the line I then turn it into a triangle, and look at the change in y, and the change in x. Finally I use the formula:

GRADIENT= VERTICAL (CHANGE IN Y AXIS)

HORIZONTAL (CHANGE IN X AXIS)

Therefore the gradient of the curve equals:

The only problem with the tangent method, is that it is only an estimation.

Increment Method: The increment method is where you plot two points (P and Q) on the curve, and the draw a line to join them. This

Middle

This is the graph of y=x5. I will continue to investigate the gradients of the graphs y=xn.

Increment Method: x=3

P(3,35) and Q(4,45)

=45-35

4-3

=781

P(3,35) and Q(3.75,3.755)

=3.755-35

3.75-3

=664.8

P(3,35) and Q(3.5,3.55)

=3.55-35

3.5-3

=564.4

P(3,35) and Q(3.25,3.255)

=3.255-35

3.25-3

=478.4

P(3,35) and Q(3.01,3.015)

=3.015-35

3.01-3

=407.7

P(3,35) and Q(3.001,3.0015)

=3.0015-35

3.001-3

=405.3

P(3,35) and Q(3.0001,3.00015)

=3.00015-35

3.0001-3

=405.03

From these results we can conclude that the gradient on an y=x, when x=3 is 405.

Instead of using general proof to work out the equation for this graph, I will use another method, that I have taught myself. It is called Binomial Expansion.

A binomial expansion is the result of a binomial expression, like (x+h) being raised to a power.

The simplest binomial expansion is (x+1), but I will show you the expansion of (x+h)5. So the expansion is:

1 x5 + 5 x h + 10 x h + 10 x h + 5 x h + 1 h

x5 + 5x4h + 10x3h2 + 10x2h2 + 5xh4 + h5

We can now, add the remainder of the formula and simplify this down, like in general proof:

x5+5x4h+10x3h2+10x2h2+5xh4+h5-x5

h

=h(5x4+10x3h+10x2h+5xh3+h4)

h

=(5x4+10x3h+10x2h+5xh3+h4)

=5x4

Therefore when:

x=1, the gradient=5

x=2, the gradient=80

Conclusion

For example, we shall take the graph, 2x²+3x³:

As we can see, two terms involved, therefore if we were to use our formula naxn-1, it would not work, as there is more than one term. Therefore we have to add something to the formula, to include the second term. So when there are formulas that have more than one term involved, the formula is:

naxn-1+laxn-1

Here the change is, the inclusion of laxn-1. This of course is the second term. The formula would, change depending on the amount of terms involved.

So therefore form this formula I can work out the gradient of the graph y=2x²+3x³.

Let’s first take the graph 2x². I, previously got the formula 4x. The second graph 3x³, we haven’t looked at before, but with our formula, we can see that it is equal 9x².

This means that to find the gradient of the curve y=2x²+3x³, we use the formula 4x+9x².

Therefore the gradient of the curve when:

x=1 is 4+9=13

x=2 is 8+36=44

x=3 is 12+81=93

x=4 is 16+144=160

If I were to work out a similar graph to, y=2x²+3x³, using differentiation, I would use exactly the same method. I would take each term in turn, and work out their respective formula, thus enabling me to put this into the general formula.

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

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