2010 Paper 1 Q5 - curve sketching of absolute functions

(i) Sketch the curve of y = |9 – x²| for  -5 ≤ x ≤ 5.

(ii)  Find the x-coordinates of the points of intersection of the curve y = |9 – x²| and the line y = 27.

This is a fundamental Sec 3 curve sketching question. It is usually paired up with quadratic functions so that we will have to do some completing the square. That’s right folks…completing the square so that we can find the turning point. 

Fortunately, in this case, we do not need to complete the square because there are roots for the quadratic function and we can simply calculate the turning point.

So, let’s get on with it…

(i)  Let y = 9 – x²

Therefore,  y = (3 + x)(3 – x). Hence, the roots are x = 3 and -3.

The turning point will occur midpoint between -3 and 3, and that would be at x = 0.

Hence, when x = 0, y = 9.

When x = 5, then y = 9 – 25 = - 16

But for the absolute value function of y, then when x = 5 or -5, y = 16.

With all these information, we will now be able to sketch the curve of y = |9 – x²| for  -5 ≤ x ≤ 5.

(ii)  To find the x – coordinates of the intersection of the curve with y = 27, we simply let y = - 27 for the quadratic function, solve for x. Why y= -27? Because, remember the function is an absolute value function. The portion of the curve below the x-axis got reflected up. Since for y = 27, it would actually have been in the y = -27 if it had not been reflected up. Hope you understood that. If not, just drop me a line again.

So,  to solve:

                                9 – x² = -27

                                x² = 36

                                x = ± 6