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

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

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.