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Level 2 Calculus NCEA 2019. Question 3c

Could you please explain the working to this question?

There are two points, A and B, on the graph of the function f (x)= x3 - 3 x2- 4 x where the tangent to the graph passes through the origin.
Find the coordinates of points A and B and the equation of each tangent.

Hi again @JOLM

This was a particularly tricky excellence question! I will draw a diagram to help explain this to you and I can post my working tomorrow for you :blush:

Thanks for your question

That was indeed a tricky question!

First of all, you need to differentiate the equation so you can find the equation of the gradient (of the tangent lines) for this graph:

f’(x)=3x^2-6x-4.

Secondly, you start deliberating: as our tangent lines pass through the origin (0,0), you can conclude that the equations of these lines should be of the form y=kx where k is a gradient of these tangent lines. Therefore k=y/x. Indeed, the gradient of any line which goes through the origin can be calculated as the y-coordinate divided by the x-coordinate of any point of the line.

Now, we know that the lines are tangent lines to our equation y=x^3 - 3x^2 - 4x. They touch the graph of this function at the points A and B. So the coordinates of these points can be written as

(x, y=x^3 - 3x^2 - 4x). Therefore:

k = y/x = (x^3 - 3x^2 - 4x)/x = x^2 - 3x - 4

From differentiation you also know that the gradient (of the tangents):

k = f’(x) = 3x^2-6x-4

So you can conclude that x^2 - 3x - 4 = 3x^2 - 6x - 4

Solving this quadratic equation you find that:

x = 0
x = 1.5

These are the x-coordinates of the points A and B. To find the y-coordinates of these points you have to substitute these values into the original equation, so you will have two points: (0,0) and (1.5, -9.375).

Lastly, you need to find the equations of these tangent lines. You already know that they can be described using the equation y=kx where k is the gradient. To find the gradient of the tangent lines at the points A and B you substitute the x-coordinates of these points into the differentiated equation:

f’(0)= -4 so the tangent line is y = -4x

f’(1.5) = -6.25 so the tangent line is y = -6.25x

Below is the graph of this function with both tangents:

image

I focused on the main steps, skipping some calculations. If you have any questions feel free to ask.

Thanks for the reply @Freydis :grinning:

For alternative working @JOLM I took a slightly different approach to this question. Firstly starting with a bit of a sketch to get my head around what is going on.

So we know the function is a cubic and that the two tangents pass through the origin.
Lets say that the x coordinate of point A is the letter “a”, the y coordinate would then be f(a) = a^3 - 3a^2-4a

There are two ways we can find the gradient of the tangent at A either using calculus or coordinate geometry methods:

  • Using calculus we differentiate to get6

  • Using coordinate geometry we could work out the rise over run:3

Both gradients should be the same as they are both for the tangent equation. We can set these equal to one another to find the value for a
4

NOTE we are getting two values for “a” as remember there are two tangents that are going through the origin. We have found both of the x values that satisfy this.

So point A is (0,0) and point B is (3/2, -75/8) where the y value is found by substitution into f(x)

From here we need to find the gradient and equations of the tangent lines. I have substituted these x valued into f’(x) to find the gradient for each point. Then used the point gradient equation to get the equation of each tangent. Working is below:

As you can see this question is well worth the excellence grade as there is a bit to do!

Thank you for your detailed answer. It is very helpful and much appreciated.

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