# A parabolla passes through the points (1,0), (2,2), (3,10). Deterine its equation.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

The equation of a parabola is a second degree equation in x. We assume y(x)= ax^2+bx+c.

Since the points (1,0), (2,2) and (3,10) are on the parabola ,  the coordinates should satisfy the equation of the parabola y(x) = ax^2+bx+c.

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

y(2) = a*2^2+b*2+c = 4a+2b+c = 2............(2)

y(3) = a*3^2+b*3+c = 9a+3b+c = 10..........(3)

(2)-(1):

(4a+2b+c)-(a+b+c) =2-0

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

(3)-(2):

(9a+3b+c)-(4a+2b+c) = 10-2

5a+b = 8...............(5)

(5) -(4):

5a-3a = 8-2 = 6

2a = 6, a = 3.

Put a = 3, in (5): 5*3 + b = 8. b = 8-15 = -7.

Put a = 3, b= -7  in (1):

3 -7 +c = 0

c = 7-3 = 4.

Therefore a = 3, b= -7 and c = 4.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

A point is on a graph if it's coordinates verify the equation of the graph.

We'll use this rule to determine the equationof the parabola.

We'll write the standard equation of a parabola:

y = ax^2 +bx + c

Since (1,0) is 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 (2,2) is on the parabola, we'll have:

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

Since (3,10) is on the parabola, we'll have:

9a + 3b + c = 10 (3)

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

4a + 2b + c - a - b - c = 2-0

We'll combine like terms:

3a + b = 2 (4)

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

9a + 3b + c - a - b - c = 10 - 0

We'll combine like terms:

8a + 2b = 10 (5)

We'll multiply (4) by -2:

-6a - 2b = -4 (6)

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

8a + 2b - 6a - 2b = 10 - 4

We'll combine and eliminate like terms:

2a = 6

We'll divide by 2:

a = 3

We'll substitute a in (4):

9 + b = 2 (4)

We'll subtract 9 both sideS:

b = 2 - 9

b = -7

We'll substitute a and b in (1):

a + b + c = 0

3 - 7 + c = 0

-4 + c = 0

c = 4

The equation of the parabola is:

y = 3x^2 - 7x + 4

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

First we start with the general equation of a parabola which is given as y = ax^2 + bx + c

Now it is given that the parabola passes through the following points:

(1 ,0)

=> 0 = a*1^2 + b*1 +c

=> a + b + c =0

(2, 2)

=> 2 = a*2^2 + b*2 +c

=> 2 = 4a + 2b +c

(3, 10)

=> 10 = a*3^2 + b*3 +c

=> 10 = 9a + 3b + c

Now using the three equations:

9a + 3b + c = 10 ...(1)

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

a + b + c =0  ... (3)

(1) - (2)

=> 5a + b =8

(2) - (3)

=> 3a + b = 2

Therefore 5a + b - 3a - b = 8-2

=> 2a = 6

=> a = 3

Substitute in 3a + b = 2

=> b= - 9 +2

=> b = -7

Substitute in a + b + c =0

=> c = -3 + 7 = 4

Therefore the parabola is:

y = 3x^2 - 7x +4

The required equation is y = 3x^2 - 7x +4

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