It’s four more days to the O-Level maths papers, and the *joss sticks* sessions, phone calls, and the churning of *customized* notes, worksheets and 小李飞刀 questions continue unabated.

All these wholesome activities have, unfortunately, combined to take an unwholesome toll on Miss Loi’s well-being, as evident from the occasional misaligned ~~fake eyelashes~~ eyelash extensions and the sometimes uneven patches of foundation on her face.

What cannot be forgiven though, was that pair of bushy, Susan Boyle-lookalike eyebrows she chanced upon in the washroom mirror this evening, which meant that she couldn’t wait for her last *joss sticks* session to end so that she could visit her beautician to restore those eyebrows to their Cindy Crawford glory again.

Unfortunately, her regular beautician had long closed for the night and she was left scouring a nearby mall for a professional who could perform this emergency corrective procedure.

Imagine her joy when she found a shop that was still open, where she gleefully ran inside to be its last customer of the day (and consequently delaying its closing time by another 10 mins).

Browsing through the glowing catalogue of eyebrow celebs that included Angelina Jolie, Ayumi Hamasaki, Sonija Kwok, and countless other unknown models, Miss Loi decided that she wanted to be Athena Chu and proceeded to lean back, close her eyes and handed over her pair of eyebrows to the ‘eyebrow stylist’ for that value-for-money $10 Athena job.

And so the stylist proceeded to *pluck* … *pluck* … *pluck* … in earnest. So rhythmic and therapeutic was her *pluck* … *pluck* … *plucking* … that Miss Loi began to slowly fall asleep, but not before seeing a familiar figure floating out of the corner of her eye … and then …

*BOOMZ!!!*

Sorry Miss! Something hit me on the head and my contact lenses fell off! Now I can’t see!

Huh? Is there anyone who can take over?

Sorry Miss! The shop is closing and I’m the only one left. You need to guide me to finish up the job!

Miss Loi then realized, to her dismay, that there were no mirrors in the shop …

With her intimate self-knowledge, however, Miss Loi knows that her eyebrow is represented by the curve `y` = −`x`^{2} + 3`x` + 5, beginning where it cuts the `y`-axis at `y` = `a` and has its arch at `x` = `b` where the maximum value of `y` is `c`.

As the eyebrow stylist had never studied nor understood calculus before, help her determine the coordinates of the arch and the starting point of Miss Loi’s eyebrow via factorising `y` = −`x`^{2} + 3`x` + 5 using the **completing the square method** to find the values of `a`, `b` and `c`.

The stylist had this to add

If you can’t guide me then I’ll have to call for the only eyebrow plucker who’s available at this hour!

## 3 Comments

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日

y = −x

^{2}+ 3x + 5 = - (x^{2}- 3x) + 5 = - (x^{2}- 2(1.5)x + (1.5)^{2}) + 5 + (1.5)^{2}= - (x - 1.5)^{2}+ 7.25Is this enough to tide you over miss loi?

cuts y-axis at y = 5, arch at x = 1.5, max value of y is at y=7.25.

曜

日

wah u stayed up til 12:58 to be the first to solve =X proness..

But u misread the most important bit, it should be a=5 ,b=1.5 ,c= 7.25. Must ans to the question...teachers minus marks for not answering to the point.(recalls blood soaked test paper)

曜

日

There are two main instances in the new Emaths syllabus where students would need to invoke the mouthful that's the completing the square method i.e. when

ax^{2}+bx+ac=0y=ax^{2}+bx+acinto a graph of the form:y= ±(x−p)^{2}+qThough the two cases look similar to the untrained eye, the perennial sin of treating the

equationax^{2}+bx+ac=0 and thefunctiony=ax^{2}+bx+acas the same thing has brought many students to grief (and blood-soaked test papers), especially after theyanyhowbring terms left and right of the = sign when dealing with a function.Obviously we're dealing with the second instant here as we need to find

a,bandcin order to save Miss Loi's eyebrow, and we'll now attempt to dissectRon's late night solution in a series ofrobuststeps:Step 1:Bring out coefficient ofx^{2}and put thexandx^{2}terms inside brackets:y= −(x^{2}− 3x) + 5The coefficient of

x^{2}is −1, so bringing it out will change the + 3xto − 3xinside the bracket (DUH!- yes simple algebra but then ...)Step 2:+ (half ofx-coefficient)^{2}− (x-coefficient)^{2}withinthe bracketsy= −(x^{2}− 3x+ (3/2)^{2}− (3/2)^{2}) + 5Step 3:Move the newminus(−)termcarefullyout of the brackets, and be mindful of the need to multiply it with an constant outside the brackets.y= −(x^{2}− 3x+ (3/2)^{2}) + (3/2)^{2}+ 5Note that the −(3/2)

^{2}has become a + (3/2)^{2}because of the − sign outside the brackets.Tread carefully here, for this is careless mistake territory.

Step 4:Transform everything inside the brackets into a square term. Whatever constant that's outside the brackets, STAYS outside the brackets.y= −(x− (3/2))^{2}+ (3/2)^{2}+ 5Note that its (

x− (3/2))^{2}NOT (x+ (3/2))^{2}because of the negative − 3xwithin the brackets.You're deep, deep into careless mistakes territory now ...

Step 5:Sum up the constant terms outside the brackets to obtain the final expressiony= ±(x−p)^{2}+q, where (p,q) are the coordinates of the:x−p)^{2}+q(Recall the happy ∪ quadratic curve)x−p)^{2}+q(Recall the sad ∩ quadratic curve )y= −(x− (3/2))^{2}+ 7.25Miss Loi's eyebrow - fierce isn't it?

And so from our final expression above, the

maximum(since there's a -ve sign outside the brackets) point of the graph is (3/2, 7.25)⇒

c= 7.25⇒

b= 3/2 = 1.5To find the

y-intercepta, simply equatexto 0 to geta=5.N.B. Performing the 5 steps above is completely safe whether you're completing the square for a function or an equation, as there's no moving of terms left and right of the = sign.