# The points A(1,1), B(0,2), C(-2,0) are on the graph of f(x)=ax^2+bx+c. Determine f?

hala718 | Certified Educator

f(x) = ax^2 + bx + c

The point A(1,1), B(0,2), and C(-2, 0) are on the graph.

Then the point A, B, and C verifies the equations:

f(1)=1

==> a+b+c= 1....(1)

f(0)= 2

==> 0+0+c= 2

f(-2)=0

==> 4a-2b + c=0.....(2)

Rewrite (1) and (2) substituting with C= 2

a+b+2 =1....(1)

4a-2b+2 =0....(2)

Now multiply (1) by 2 and add to (2)

==> 6a +6 = 2

==> 6a = -4

==> a= -4/6= -2/3

==> b= -1-a= -1+2/3= -1/3

Then the equation is :

f(x) = (-2/3)x^2 -(1/3)b + 2

To check:

f(1) = -2/3 -1/3 +2 = 1

f(0)= 2

f(-2) = -8/3 +2/3 +2 =-6/3 +2= -2+2=0

giorgiana1976 | Student

If the points are on the graph of the function, that means that their coordinates, verify the equation of the function:

A belongs to f, so:

f(1)=1

a+b+c=1

B belongs to f, so:

f(0)=2

c=2

C belongs to f, so:

f(-2)=0

4a-2b+c=0

But c=2, so 4a-2b=-2 (1)

and

a+b=-1 (2)

We'll multiply (2) by 2 and we'll add it to (1):

4a-2b+2a+2b=-2-2

6a = -4

a = -4/6

a = -2/3

-2/3+b=-1

b = -1 + 2/3

b = -1/3

The function is:

f(x) = (-2/3)x^2 + (-1/3)x + 2

neela | Student

We substitute the coordinates of the points A(1,1),B(-2,0), C(-2,0)in the equation ax^2+bx+c  = y.

a*1+b*1+c = 1...........(1)

a*0+b*0+c = 2..........(2). Threfore c = 2.

a*(-2)^2+b*(-2) +c = 0.....(3)

4a-2b +c = 0 , 4a-2b= -c = -2,  2a-b = -1.....(4)

From (1) a+b = 1-c= 1-2 = -1. Or

a+b =-1....(5)

Eq (4) + Eq(5) :  3a = -2, a = -2/3

So  from (1) = a+b+c =-1,  (-2/3)+b+ 2 = -1, b = 1-2+2/3 = -1/3.

Therefore ax^2+bx+c = y is

-2/3x^2-1/3 x +2 =y