I have a point $Q$ that lies on the curve $C$, where $C=\frac{3x^2}{4}-4x-10$. The point $Q$ is such that the gradient of the normal to $C$ at $Q$ is $-2$, how can I find the $x$-coord of $Q$?
So far I have tried differentiation however I am not very good at it since the fraction coefficient confuses me!
So first we will look at how to differentiate $C$. Don't worry about the fractional coefficient we can take that out if you'd like. So:
$$\frac{3x^2}{4}=\frac{3}{4}(x^2) \text{ then differentiating} \rightarrow \frac{3}{4}(2x)=\frac{3x}{2}$$
So if we differentiate $C$ we get $C'=\frac{3x}{2}-4$. Which is the gradient at any point $x$. Also what we should know is the tangent is the gradient at that point. But the normal is perpendicular to it! Which means in terms of the tangent it is the negative reciprocal ! So if our tangent's gradient was $2$, then the normals gradient would be $\frac{-1}{2}$.
So generally:
$$\text{Gradient of Tangent}=\frac{-1}{\text{Gradient of Normal}}$$
So we have the gradient of the normal which is $-2$, so finding the gradient of the tangent we get $\frac{1}{2}$ from the method above! So then we can set our new gradient equal to our $C'$ formula! $$C'=\frac{1}{2}=\frac{3x}{2}-4$$
I'm sure you can do the last step! Just re-arrange for x! Hope this helps!