2010 Paper 1 Q8 - tangents

The equation of a curve is y = x³ + 3x² – 9x + k, where k is a constant.

(i) Find the set of values of x for which y is decreasing.

(ii) Find the possible values of k for which the x-axis is a tangent to the curve.

Hmm….it seems that there are quite a bit of differential questions in this paper.  I guess that is good. We may be tending towards the earlier years when I was taking the A maths when more than 50% was calculus.

Anyway, let’s discuss part (i). When is it that the function is increasing, or when the function is decreasing?

Well, we return to dy/dx!  Again, what does dy/dx mean? It means the change in y with respect to a change in x. If a change in x creates a positive change in y, it means that y is increasing with every change in x.   When a change in x leads to a negative change in y, that means that y is decreasing.  Hence, when we want to find the set of values of x for which y  is decreasing, then we are going to find when dy/dx < 0.

Given that y = x³ + 3x² – 9x + k, then

For decreasing values of y, then let dy/dx < 0

Hence the solution set is  {x: x Є R,  - 3 < x < 1 }

Again, don't forget to put it in set notation because the question asked for the solution set.

(ii) To find the possible values of k, (and in this case we know that there will be more than one value of k since the question asked for possible values) , we need to remember that k is a constant. In this case, the only effect that k has in the curve is to move it up or down. Hence, what we want to know is, which value of x will produce a gradient of zero (parallel to the x-axis) and then from there, put in the value of x into the equation for y, and then equating it to zero, and we will find the value of k. Confused? Nevermind. Let’s see it in action. That is clearer.

When dy/dx = 0, then  x = 1 or x = -3

Substitute x = 1 into y  and equating that to zero brings:

1³ + 3(1)² – 9(1) + k = 0

1 + 3 – 9 + k = 0

k = 5

Substitute x = -3 into y and equating that to zero gives:

(-3)³ + 3(-3)² – 9(-3) + k = 0

-27 + 27 + 27 + k = 0

k = -27

Hence, when k = 5 or k = -27, the x-axis is a tangent to the curve.