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

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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**