## Find the quadratic function that passes through the points: (-1,6), (1,4), and (2,9). y=ax²+bx+c

Question

Find the quadratic function that passes through the points: (-1,6), (1,4), and (2,9).
y=ax²+bx+c

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6 days 2021-09-12T06:10:53+00:00 1 Answer 0

## Answers ( )

The quadratic function that passes through given points is                           y = 2 x² – x + 3  .

Step-by-step explanation:

The given quadratic function as

y = a x² + b x + c

The equation passes through the points ( – 1 , 6 ) , ( 1 , 4 ) and ( 2, 9 )

As The points passes through equation then

At points  ( – 1 , 6 )

6 = a (1)² + b ×( – 1 ) + c

Or, a – b + c = 6           …..A

Again At points  ( 1 , 4 )

4 = a (1)² + b × 1 + c

Or, a + b + c = 4            …….B

Similarly At points  ( 2 , 9 )

9 = a (2)² + b × 2 + c

Or, 4 a +2 b + c = 9        ….,,,C

Now solving equation A and B

(  a – b + c ) + (  a + b + c ) = 6 + 4

Or, a + c =

I.e a + c = 5           ……D

Similarly Solving equation B and C

( 4 a +2 b + c  ) – 2 × ( a + b + c ) = 9 – 2 × 4

Or, ( 4 a – 2 a + 2 b – 2 b + c – 2 c ) = 9 – 8

Or, ( 2 a – c ) = 1        …..E

Solving D and E

( a + c ) + ( 2 a – c ) = 5 + 1

Or, 3 a = 6

∴  a =

I.e a = 2

Put the value of a in Eq D

So ,  a + c = 5

Or,  c = 5 – a

∴  c = 5 – 2 = 3

I.e  c = 3

Put The value of a and c in Eq A

a – b + c = 6

Or, b = a + c – 6

Or . b = 2 + 3 – 6

∴ , b = 5 – 6

I.e   b = – 1

Now, Putting the values of a , b , c in the given quadratic equation

I.e y = a x² + b x + c

Or, y = 2 x² + ( – 1 ) x + 3

∴ The quadratic eq is  y = 2 x² – x + 3

Hence The quadratic function that passes through given points is                 y = 2 x² – x + 3  . Answer