How can we find the gradient of the line through the points with coordinates \((3,9)\) and \((3.5,12.25)\)?

How can we find the gradient of the line through the points with coordinates \((3,9)\) and \((3.1,9.61)\)?

How can we find the gradient of the line through the points with coordinates \((3,9)\) and \((3.01,9.0601)\)?

To find the gradient of the line going through the points \((x_1,y_1)\), \((x_2,y_2)\) we look at the change in the \(y\)-coordinate divided by the change in the \(x\)-coordinate:

\[ \frac{y_2-y_1}{x_2-x_1}.\]

So for each of the pairs of points we have:

\(\dfrac{12.25-9}{3.5-3}=6.5\).

\(\dfrac{9.61-9}{3.1-3}=6.1\).

\(\dfrac{9.0601-9}{3.01-3}=6.01\).

Can we make any suggestions on what is happening?

What could we suggest the gradient of the tangent line at \((3,9)\) to the curve \(y=x^2\) is?

How can we generalise this for any point \((x,x^2)\)?

Can you use a spreadsheet to investigate what the tangent line gradient will be for \(x=2\), \(x=4\) etc…?

How can we find the gradient of the line through the points with coordinates \((x,x^2)\) and \((x+h,(x+h)^2)\)?

We can apply the same thoughts when we had numbers,

So, the gradient of the line through the points \((x,x^2)\) and \((x+h,(x+h)^2)\) is \(2x+h\).

As we move the point \((x+h,(x+h)^2)\) closer and closer to \((x,x^2)\), we are making \(h\) smaller and smaller, finally making \(h\) zero. This gives the gradient of the line that touches the curve \(y=x^2\) at the point \((x,x^2)\) as \(2x\).

Looking at the cubic \(y=x^3\)

Can you use a spreadsheet to investigate what the tangent line gradient will be for \(x=1\), \(x=2\), \(x=3\) etc… on the cubic curve \(y=x^3\)?

If we take \(x=2\), we have the point \((2,8)\) and we can choose a second close point on the curve \(y=x^3\), \((2.01,9.261)\). This would give a gradient of

So, the gradient of the line through the points \((x,x^3)\) and \((x+h,(x+h)^3)\) is \(3x^2+3hx+h\).

As we move the point \((x+h,(x+h)^3)\) closer and closer to \((x,x^3)\), we are making \(h\) smaller and smaller, finally making \(h\) zero. This gives the gradient of the line that touches the curve \(y=x^3\) at the point \((x,x^3)\) as \(3x^2\).