Matrices Question (1 Viewer)

[ScottyG]

I don't do 4U, this is for a friend who needs it desperately for an assessment task. He's doing the VCE, but I was told that Matrices exist in the 4U Maths HSC course.

If so:

Using a matrix technique find, in terms of k, the function g( x ) which represents the faimily of the quadratic graphes that go through the points A (1,4), B (2,2) C(0,k)

I top out at 2U, so it's about 4 leagues above me.

 

[LoneShadow]

let y = ax²+bx+c
so for A(1,4):
4 = a+b+c
for B(2,2):
2 = 4a+2b+c
for C(0,k):
k = c

for a matrix:

1 1 1 | 4
4 2 1 | 2
0 0 1 | k

reduce it and you'll get a = (k/2 - 3); b = (7 - 1.5k); c = k

so y = (k/2 - 3)x² + (7 - 1.5k)x + k

 

[ScottyG]

Looks like it's right. Cheers LoneShadow.

 

[Riviet]

Just to clarify, it's to my knowledge that matrices are not in any HSC mathematics course, they are explicitly encountered in uni.

 

[adgala]

i remember doing matrices back in year 8 or 9 but only very simple concepts and nothing like the above question

 

[ScottyG]

Your success has prompted more questions, lol.



It's from a past VCE test, he and I have no idea.

 

[Templar]

b and c are quite obvious once you play around with values of k.

For d, y = (k/2 - 3)x² + (7 - 1.5k)x + k
y'=(k-6)x+(7-3k/2)

For e, turning points ie y'=0
x=(3k-14)/(2k-12) if k-6 is not zero. Otherwise it's a linear equation instead of a quadratic.

2a, we can deduce the following equations from y=ax²+bx+c
1.5=16a+4b+c
2=4a+2b+c

In addition, at point (2,2), y'=2k-12+7-3k/2=k/2+5
k/2+5=4a+2b

Matrix:
16 4 1 | 1.5
4 2 1 | 2
4 2 0 | k/2+5

Solving it we get
a=-1/16(11+k)
b=1/8(31+k)
c=-1/2(k+6)

Part b is just a trivial substitution and check it works.

 

[Roobs]

hmm..just for the record Matricies are not in NSW 4unit, but i know that Queensland does them