# Find the quadratic whose graph is passing through the points (-1,3), (0,-1), (2,4)

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### 3 Answers

Let's start with the general equation of a quadratic equation: y = ax^2+ bx +c.

Now this passes though the points (-1,3), (0,-1), (2,4).

So we have:

- 3 = a*(-1)^2+ b*(-1) +c = 0.

=> 3 = a - b +c

- -1 = a*0 + b*0 +c

=> -1 = c

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

=> 4 = 4a + 2b +c

Now we can substitute c= -1 in 3 = a - b +c and 4 = 4a + 2b +c to get the equations

3= a - b -1 and 4 = 4a + 2b -1

=> 4 = a - b and 5 = 4a + 2b

=> 16 = 4a - 4b and 5 = 4a + 2b

From the above we get 16 - 5 = 4a - 4b - 4a - 2b

=> 11 = -6b, so b = -11/6

Now 4 = a - b => a = b + 4 = -11/6 + 4 = (24 -11 )/6 = 13/6

**Therefore the quadratic function is **

**y = (13/6)x^2 - (11/6)x -1**

We assume the y = ax^2+bx+c passes through (-1, 3), (0-1) and (2,4).

Therefore

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

3 = a-b +c..............................(1)

-1 = a(0)^2+b(0) +c , so -1 = c . Or c = -1.

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

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

Eq (1) *2 + Eq(2) gives :

3*2+4 = 2a+4a-2b+2b +2c+c

10 = 6a+3 c = -1.

10 = 6a -3

6a = 10+3. Or a = 13/6

Therefore from (1) , b = a+c-3 = 13/6-1-3 = -11/6

Therefore y = (13/6)x^2-(11/6)x-1

We'll write the general form of the quadratic:

ax^2 + bx + c = y

If the graph passes through the given points, that means that the coordinates of the points verify the equation of the quadratic.

The point A(-1,3) belongs to the graph if and only if

yA = axA^2 + bxA + c

We'll substitute the coordinates of the point A into the equation:

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

a - b + c = 3 (1)

The point B(0,-1) belongs to the graph if and only if

yB = axB^2 + bxB + c

We'll substitute the coordinates of the point B into the equation:

**c =- 1**

The point C(2,4) belongs to the graph if and only if

yA = axC^2 + bxC + c

We'll substitute the coordinates of the point C into the equation:

4 = 4a + 2b - 1

We'll add 1 boh sides:

4a + 2b = 5 (2)

We'll substitute the value of c in (1):

a - b -1 = 3 (1)

We'll add 1 both sides:

a - b = 4 (3)

We'll multiply (3) by 2:

2a - 2b = 8 (4)

We'll add (4)+(2):

4a + 2b + 2a - 2b = 5+8

6a = 13

We'll divide by 6:

**a = 13/6**

We'll substitute a in (3):

13/6 - b = 4

We'll subtract 13/6 both sides:

-b = 4 - 13/6

-b = 11/6

**b = -11/6**

The quadratic is:

**f(x) = (13/6)*x^2 - (11/6)*x - 1**