1. ## help with finding constants in quadratice function

This is the problem I'm having trouble with right now and its racking my brain, any help would be appreciated:

Consider a quadratic function Q(y)=A⁢y2+B⁢y+C. If Q(2)=0, Q(3)=0 and Q(1)=-4, find C the constant term in Q.

2. Originally Posted by forkball42
This is the problem I'm having trouble with right now and its racking my brain, any help would be appreciated:

Consider a quadratic function Q(y)=A⁢y2+B⁢y+C. If Q(2)=0, Q(3)=0 and Q(1)=-4, find C the constant term in Q.
What are those boxes (⁢)? Your question appears is appearing as Q(y)=A⁢y2+B⁢y+C. Is it $Q(y) = Ay^2 + By + C$?

For this question, substitute $y=2, 3, 1$ and you will get three equation which you can use simultaneous equation technique to work out constant value of $A, B$ and $C$.

3. Originally Posted by forkball42
This is the problem I'm having trouble with right now and its racking my brain, any help would be appreciated:

Consider a quadratic function Q(y)=A⁢y2+B⁢y+C. If Q(2)=0, Q(3)=0 and Q(1)=-4, find C the constant term in Q.
$Q(y)=Ay^2+By+C$

for Q(2)=0,

$0=A(2)^2+B(2)+C$

4A + 2B + C =0 .......................................eqn. (1)

for Q(3)=0,

9A + 3B + C = 0 ..................................... eqn.(2)

for Q(1) = - 4,

A + B + C = - 4 .......................................eqn.(3)

subtract eqn(3) from eqn(1) and eqn (2), we get:

3A + B = 4 ............................................EQN(4)

and 8A + 2B =4
or,
4A + B =2 ..........................................EQN(5)

Solve EQN(4) AND EQN(5):
we get:
A = - 2, B = 10
Put these values in eqn(3), we got C = - 12