Review question

# What is the gradient of this curve at its $x$-intercepts? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7391

## Solution

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?