# equation/functionFind the equation of the quadratic function with the points (-1,4), (1, -2) and (2,1)

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You need to determine the quadratic function `y = ax^2 + bx + c, ` under the given conditions, hence, you need to solve the following system of equations, such that:

`{(a - b + c = 4),(a + b + c = -2),(4a + 2b + c = 1)}`

You need to evaluate determinant of matrix of system, such that:

`Delta = [(1,-1,1),(1,1,1),(4,2,1)]`

`Delta = 1 + 2 - 4 - 4 - 2 + 1 = -6`

`a = Delta_a/Delta`

`Delta_a = [(4,-1,1),(-2,1,1),(1,2,1)]`

`Delta_a = 4 - 4 - 1 - 1 - 8 - 2 = -12`

`a = (-12)/(-6) => a = 2`

`b = Delta_b/Delta`

`Delta_b = [(1,4,1),(1,-2,1),(4,1,1)]`

`Delta_b = -2 + 1 + 16 + 8 - 1 - 4 = 18`

`b = 18/(-6) = -3`

Replacing 2 for a and -3 for b in the middle equation `a + b + c = -2 ` yields:

`2 - 3 + c = -2 => c = 3 - 4 => c = -1`

**Hence, the quadratic function is fully determined, under the given conditions, for **`a = 2, b = -3, c = -1.`

We'll write the equation of the parabola:

y = ax^2 + bx + c

Since the parabola is passing through the points (-1,4), (1, -2) and (2,1), we'll have:

f(-1) = 4

We'll substitute x by -1 in the expression of quadratic:

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

a - b + c = 4 (1)

f(1) = -2

We'll substitute x by 1 in the expression of quadratic:

a(1)^2 + b(1) + c = -2

a + b + c = -2 (2)

f(2) = 1

We'll substitute x by 2 in the expression of quadratic:

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

4a + 2b + c = 1 (3)

We'll add (1) + (2):

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

We'll combine and eliminate like terms:

2a + 2c = 2

We'll divide by 2:

a + c = 1 (4)

We'll add (3) + 2*(1):

4a + 2b + c + 2a - 2b + 2c = 8 + 1

We'll combine and eliminate like terms:

6a + 3c = 9

We'll divide by 3:

2a + c = 3 (5)

We'll subtract (4) from (5):

2a + c - a - c = 3 - 1

We'll combine and eliminate like terms:

a = 2

4 + c = 3

c = 3 - 4

c = -1

a - b + c = 4

2 - b - 1 = 4

-b = 4 - 2 + 1

b = -3

The quadratic equation is:

y = 2x^2 - 3x - 1