The information on top of the diagram is always important, give it a shot and take something out of it. From this one we can add point B on the cartesain plane, the rest is on the diagram and we are good. At first, I thought point B is on the derivative, the parabola, but it’s actually on the main graph, the cubic function f, of which is not drawn above.

Note:

The cubic function has a first derivative which is a parabolic function

The first derivative of a function describes gradient at any point

We usually use the first derivative to find gradient of a tangent

8.1

We have all the information necessary to determine the equation of the parabola above

The first thing I thought of was to use the turning point formula and I went with that

Substitute the turning point on p and q

P is the x value of TP and changes sign from the cartesain plane to the equation and vice versa

Q is the y value of TP as it is

Substitute any other point to find the value of a, I personally chose the x intercept (-1;0)

8.2

This is a pretty new question, right?

Well, at least to me

I didn’t know what to do, that ‘determine t’ confused the hack out of me. But I kept my faith and knew that I have to multiply the entire first derivative by ½, and I did just that.

To my surprise, that’s what they wanted

On both 8.2.1 and 8.2.2 we do the same thing, adjust as they want, see on answer sheet below

8.3

The turning points of the cubic function are share the x values of the x intercepts of the parabola, the first derivative

So, it’s simply negative 5 and negative 1 in this case

8.4

Let’s break it down

We can find the gradient of a tangent neh, using the first derivative as it describes gradient at a point

And the we can find the gradient of the perpendicular line k, using product of gradients.

And the we use point B to find the value of c as in the answer sheet below

Answer: