maths

P:y=x^2+bx+c A(-5,-9) and B(1,-3) y=px+q

2010-09-03 21:04:07 補充：
-9=25-5b+c
5b-c=34

-3=1+b+c
b+c=-4

5b-c+b+c=34-4
6b=30
b=5

5+c=-4
c=-9

so P: y=x^2+5x-9

2010-09-03 21:04:52 補充：
then we sub T into P
px+q=x^2+5x-9
x^2+(5-p)x+(-9-q)=0
because meet at 1 pt only
det=0
(5-p)^2-4(1)(-9-q)=0
(5-p)^2+36+4q=0

also they intersect at (2,5)
i.e. 5=2p+q
q=5-2p

(5-p)^2+36+4q=0
p^2-10p+25+36+4(5-2p)=0
p^2-10p+25+36

2010-09-03 21:06:00 補充：
p^2-10p+25+36+20-8p=0
p^2-18p+81=0
(p-9)^2=0
p=9
q=-13

P and T intersect at D(2,5)..pls help

Is some information missing??