Find the quadratic function if the parabola passes through points (1,0),(2,1) and (3,2)?

The given points lie on the parabola if the corresponding
coordinates verify the quadratic equation of the
parabola.

We'll recall the standard equation of a
parabola:

y = ax^2 +bx +
c

Since (1,0) lies on the parabola, we'll
have:

0 = a + b + c

We'll use
symmetric property and we'll get:

a + b + c = 0
(1)

Since (2,1) is on the parabola, we'll
have:

4a + 2b + c = 1
(2)

Since (3,2) is on the parabola, we'll
have:

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

We'll subtract (1) from (2) and we'll
get:

4a + 2b + c - a - b - c =
1-0

We'll combine like
terms:

3a + b = 1 (4)

We'll
subtract (1) from (3) and we'll get:

9a + 3b + c - a - b -
c = 2 - 0

We'll combine like
terms:

8a + 2b = 2 (5)

We'll
multiply (4) by -2:

-6a - 2b = -2
(6)

We'll add (6) to (5):

8a +
2b - 6a - 2b = 2 - 2

We'll combine and eliminate like
terms:

2a = 0 => a =
0

We'll substitute a in (4):

b
= 1 (4)

We'll substitute a and b in
(1):

a + b + c = 0

0 + 1 + c =
0

1 + c =
0

c=-1

Since the
coefficient of x^2 is cancelling, the quadratic function degenerates to a linear
function y=bx+c: y=x-1