In each case can you find the equation of the curve which has the given gradient function and the previously matched tangent and normal lines?

We know the gradient function for each curve and since we know the point where the tangent and normal lines meet, we know a point on the curve.

If we have a gradient function of \(6x^2\), is it possible to find the original function for the curve?

If we start with a function of \(x\) and differentiate we get the gradient function for that curve. Can we go back?

If we end up with \(6x^2\) we must have started with \(x^3\) as our power of \(x\), but if we differentiate \(x^3\) with respect to \(x\) we get \(3x^2\). Therefore, we could have started with \(2x^3\).

What happens to \(2x^3+4\) when you differentiate with respect to \(x\)?

Let us go back to our first example, pair one. We have matched up the tangent \(y+2x+1=0\) with the normal \(2y-x-3=0\) to the curve with gradient function \(\dfrac{-2}{x^2}\) at the point \((-1,1)\).

If we start with the gradient function, the function for the curve we started with must end up in this form. Since we end up with \(\dfrac{-2}{x^2}\) we should have started with \(2x^{-1}\) for our power of \(x\).

How do we know this? What happens when we differentiate \(x^{-1}\) with respect to \(x\)?

However, we know that constants when differentiated go to zero, so the most general form we could have started with for this function is \(y=\dfrac{2}{x}+c\), where is \(c\) is real number.

How can we find this constant \(c\)?

From our work with the tangent and normal lines, we know a point on the curve (the point where the tangent and normal lines meet), \((-1,1)\).

Therefore, when \(x=-1\), \(y=\dfrac{2}{(-1)}+c=1\) and so \(-2+c=1\), giving \(c=3\). This tells us that the original function we started with is \(y=\dfrac{2}{x}+3\).

We can do the same for each of the pairs of lines and their intersection points giving the original curves as:

\(y=x^2+5x+6\) for the gradient function \(2x+5\).

\(y=5x^3+6x^2+3x+2\) for the gradient function \(15x^2+12x+3\).

\(y=\dfrac{2}{x}+3\) for the gradient function \(\dfrac{-2}{x^2}\).

\(y=8x^3-4x^2+6x-5\) for the gradient function \(24x^2-8x+6\).