A familiar engine roar filled the air as the bus they had been waiting for pulled into the bay at the interchange, creating a temporary breeze that momentarily lifted the hunk’s silky flowing hair into thousands of little cosine waves.

Invoking her supreme mathematical powers honed through her years at The Temple, our cutie π schoolgirl managed to narrow down the location of the hunk’s π-spot to that π tattoo on his shoulder!

As Fate had it, the bus door hissed open that very moment and the queue began to move.

And as Fate would have it, all that stood between the hunk and the schoolgirl was one big fat ugly butch with a waistline that stretched from railing to railing, and who was rapidly moving along at a top speed of 1^{-10}m/hr.

CHIJ uniform [source] – for those who’ve never seen one

With the hunk rapidly disappearing into the distance , along with her hopes of showing him off to her bitchy Eurasian girlfriends, the schoolgirl decided to throw caution to the wind and did the most un-girly thing imaginable – she scaled the railing in her CHIJ school uniform, crudely brushed aside an insurance agent asking her to do a survey, to make a last-ditch dash for that shoulder of the hunk who was already boarding the bus.

Time seemed to slow down, Matrix-style, as she lunged herself towards that inviting π tattoo, as her future, her fate, her destiny, hinged on this moment when the bus doors began to hiss again …

Given that, as she lunged herself at the last moment, the movement of the schoolgirl’s outstretched hand followed a path represented by a quadratic curve, and that:

The quadratic curve is symmetrical about the line x = 2 and passes through the points (-1, -11) and (4, 9). Find the coordinates of the maximum or minimum point of the curve.

And given that the π tattoo is located at the coordinate (2, 26), did she make it?

IMPORTANT: In Quadratic Equations questions, whenever you see the phrase symmetrical about the line x = h (where h can be any number), that’s normally the cue for you to use your quadratic expression in the completed square form i.e.

y = a(x – h)^{2} + k

And needless to say, when a quadratic curve is symmetrical about h, it’s maximum/minimum point will occur at x = h 😉

Personally, I feel that the question will be much more interesting if just one coordinate is given 🙂 Since the graph is symmetrical about x=2, giving one coordinate is akin to giving two 🙂 So if (-1,-11) is given, then we should know that (5, -11) is another point that the curve passes through.

Anyway, the general form of this quadratic eq is: y= a (x-2)² + b because the max/min pt lies on the line of symmetry, hence the x-coordinate of the max/min pt is also 2. In fact, it's (2, b)

sub in (-1, -11) and (4,9) and we get:

-11 = 9a + b ------(1) 9 = 4a + b --------(2)

Solving the simultaneous eq, a= -4, b=25

Hence, the quadratic eq is y = -4(x-2)² + 25. Since the coefficient of x² is -ve, it must be an 'unhappy face' curve with a maximum pt, thus

max pt is (2, 25)

Since the tattoo is at (2,2526), she just missed it by 1 y-unit 🙂

Wah Papillion, you've said everything that Miss Loi wanted to say LOL

Yes you'll only really need one set of coordinates to set things going since once the line of symmetry (x = 2) is given, you should be able to find the x-corrdinate of the second set of coordinates by reflecting about x = 2 i.e.

Given (-1, -11) ------> (2+[2-(-1)], -11) = (5, 11) or Given (4, 9) -------> (2-[(4)-2], 9) = (0, 9)

And please, please remember that in your y = a(x - h)^{2} + k quadratic equation.

1. h is the x-coordinate of the line of symmetry AND the max/min point of graph (should be common sense anyway). 2. k is the y-coordinate of the max/min point of the curve (again common sense).

So simply sub-in any two coordinates to form two simultaneous equations as shown by Papillion's workings above and you'll be able to find a = -4 and k = 25.

To find whether it's a max or min point, all students should know by heart the two 'quadratic moods, regardless of whether the equation is in the form of y = a(x-h)^{2}+k or y=ax^{2}+bx+c

So

1. If a is in a positive mood, the curve smiles 🙂 ⇒ minimum point

2. If a is in a negative mood, it frowns 🙁 ⇒ maximum point

So since a is negative here, we are looking at a maximum point of y = 25, which our girl just missed by 1 unit!

So poor thing hor the girl? And now the a coefficient of the quadratic equation of her mouth must be very, very negative!

P.S. Really, Miss Loi thought of leaving only one set of coordinates for this question. But decided in the end to leave it intact (as it appeared from a particular textbook) as some of her students already find it challenging enough trying to remember all the points list ... sigh ...

* Please refer to the relevant Ten-Year Series for the questions of these suggested solutions.

About Miss Loi

Miss Loi is a full-time private tutor in Singapore specializing in O-Level Maths tuition. Her life’s calling is to eradicate the terrifying LMBFH Syndrome off the face of this planet. For over years she has been a savior to countless students … [read more]

Gratitudes

Under Ms Loi, I progressed tremendously and in my preliminary examinations a month later, I actually scored A1s for both my A and E maths … [read more]

## 3 Comments

曜

日

Hee hee, found a question that is unanswered 🙂

Personally, I feel that the question will be much more interesting if just one coordinate is given 🙂 Since the graph is symmetrical about x=2, giving one coordinate is akin to giving two 🙂 So if (-1,-11) is given, then we should know that (5, -11) is another point that the curve passes through.

Anyway, the general form of this quadratic eq is:

y= a (x-2)² + b

because the max/min pt lies on the line of symmetry, hence the x-coordinate of the max/min pt is also 2. In fact, it's (2, b)

sub in (-1, -11) and (4,9) and we get:

-11 = 9a + b ------(1)

9 = 4a + b --------(2)

Solving the simultaneous eq, a= -4, b=25

Hence, the quadratic eq is y = -4(x-2)² + 25. Since the coefficient of x² is -ve, it must be an 'unhappy face' curve with a maximum pt, thus

max pt is (2, 25)

Since the tattoo is at (2,

~~25~~26), she just missed it by 1 y-unit 🙂曜

日

Wah

Papillion, you've said everything that Miss Loi wanted to say LOLYes you'll only really need

oneset of coordinates to set things going since once theline of symmetry(x= 2) is given, you should be able to find thex-corrdinate of the second set of coordinates by reflecting aboutx= 2 i.e.Given (-1, -11) ------> (2+[2-(-1)], -11) = (5, 11) or

Given (4, 9) -------> (2-[(4)-2], 9) = (0, 9)

And please, please remember that in your

y=a(x-h)^{2}+kquadratic equation.his thex-coordinate of the line of symmetry AND the max/min point of graph (should be common sense anyway).2.

kis they-coordinate of the max/min point of the curve (again common sense).So simply sub-in any two coordinates to form two simultaneous equations as shown by

Papillion's workings above and you'll be able to finda= -4 andk= 25.To find whether it's a max or min point, all students should know by heart the two 'quadratic moods, regardless of whether the equation is in the form of

y=a(x-h)^{2}+kory=ax^{2}+bx+cSo

ais in apositivemood, the curvesmiles🙂 ⇒ minimum point2. If

ais in anegativemood, itfrowns🙁 ⇒ maximum pointSo since

aisnegativehere, we are looking at amaximumpoint ofy= 25, which our girl just missed by 1 unit!So poor thing

horthe girl? And now theacoefficient of the quadratic equation of her mouth must be very, very negative!P.S. Really, Miss Loi thought of leaving only one set of coordinates for this question. But decided in the end to leave it intact (as it appeared from a particular textbook) as some of her students

alreadyfind it challenging enough trying to remember all the points list ... sigh ...曜

日

Oh BTW just corrected that little typo of yours 🙂