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[sco11]

find quadratic equation of function that has minimum at x = 2 and goes through points (0,2) and (3,1)

 

[Guest]

quadratic eqn has the form of
y= ax^2 +bx + c

use the points (0,2) and (3,1)

2= 0 + 0 + c

2=c

1= a(3)^2 + b (3) + c
1 =9a + 3b + 2

-1 = 9a + 3b

for the minimum at x=2

dy/dx = 2ax + b

but dy/dx = 0 at this point

0= 2ax + b

b= -2ax

sub this back into -1 = 9a + 3b and solve...

all the best