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

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

Answers ( )

    0
    2021-09-12T06:11:54+00:00

    Answer:

    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 = \frac{10}{2}  

    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 = \frac{6}{3}

    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

Leave an answer

Browse
Browse

27:3+15-4x7+3-1=? ( )