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

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### 3 Answers

Let the function be :

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

Given the points (-1,4), (1, -2) and (2,1) are on the curve of f(x):

Then the points should verify the equation such that:

f(-1) = 4 f(1) = -2 and f(2) = 1

Let us substitute:

f(1) = a+ b + c = -2 .............(1)

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

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

Now we will solve the system:

First we will add ( 1) and (2):

==> 2a + 2c = 2

==> a + c = 1..............(4)

Now multiply (2) by 2 and add to (3):

==> 2a - 2b + 2c = 8

==> 6a + 4c = 9..............(5)

==> Multiply (4) by -4

==> -4a - 4c = -4

==> Now add to (5):

==> 2a = 4

**==> a= 2**

**==> c = -1**

**==> b = -3**

**==> f(x) = 2x^2 - 3x - 1**

Let ax^2+x+c = y(x) which has the points(-1,4), (1-2) and (2,1).

So we put the coordinte values in the equation:

(-1,4):

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

(1,-2):

a + b +c = -2.......(2).

(2,1):

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

(1)-(2) gives: -2b = 4-9-2) = 6. So b = 6/-2 = -3.

(1)-(3) gives: -3a -3b = 3. a+b = -1. So a = -1-b = -1-(-3) = 2.

Put a= 2 and b = -3 in (2): 2 -3 +c = -2. So c= -2-2+3 = -1.

Therefore a = 2, b = -3 and c = -1.

Therefore y = 2x^2 -3x -1 is the quadratic function that has the given 3 points on it.

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**