The gradient of the curve \(y=3x^2+5x-12\) is \(17\) at the point \(P\). Calculate the coordinates of \(P\).

The gradient of the curve is given by \[ \frac{dy}{dx}=6x+5. \] We must find the value of \(x\) for which \[ 6x+5=17. \] Subtracting \(5\) from both sides, we see that \[ 6x=12, \] and therefore \[ x=2. \] Therefore \(P=(2, 3\times 2^2+5\times 2-12)=(2,10)\).

The curve cuts the \(x\)-axis at \(Q\) and \(R\). Find the gradient of the curve at \(Q\) and at \(R\).

We must first find \(Q\) and \(R\) by finding the roots of \(y=3x^2+5x-12=0\). By inspection, we can see that \[3x^2+5x-12=(3x-4)(x+3),\] and so \[Q=(-3,0),\quad R=(\frac{4}{3},0).\]

Evaluating \(\dfrac{dy}{dx}\) at these points, we find that at Q, \[ \frac{dy}{dx}=6\times -3+5=-13, \] and at R, \[ \frac{dy}{dx}=6\times \frac{4}{3}+5=13. \] So the gradient of the curve at \(Q\) is \(-13\), and at \(R\) is \(13\).

Why do the two gradients differ only in their sign?