Find the quadratic function passing through the points (0,-3), (1,2), and (2,-1)

The quadratic function passing through the points (0,-3), (1,2), and (2,-1) is

Solution:

Given that required function is quadratic  

And function is passing through points (0 , -3) , (1 , 2) and (2 , -1)

General form of a quadratic function is   ----(A)

f(x) is nothing but output value that is y.

That is f(x) = y

So  --- (1)

Let’s use equation (1) to get required function.

Given that function passes through (0 , -3) means when x = 0 , y = -3

On substituting value of x and y in equation (1) we get  

-3 = 0 + 0 + c

c = -3

On substituting value of c in equation (1) we get  

---(2)

Function also passes through point (1, 2) that is at x = 1 , y = 2.

On substituting value of x and y in equation (2) we get  

2 = a + b – 3

a + b = 5                

b = 5 - a   -------(3)

Also given function passes through point ( 2 , -1) means when x = 2 , y = -1

On substituting value of x and y in equation (2) we get  

-1 = 4a + 2b – 3

4a + 2b = 2

2a + b = 1  ------- (4)

On substituting value of b from equation (3) in equation (4), we get

2a + (5 - a ) = 1  

a + 5 = 1

a = 1-5 = -4

From equation (3) b = 5 – a = 5 – (-4) = 9

b = 9

Now we have a = -4, b = 9 and c = -3

On substituting calculated values of a, b, and c in equation (A) we get

Hence required quadratic function is

coin flip

non-answer:

choosing

sorry if this isnt what you're looking for i dont really understand the question

step-by-step explanation: 4 cause 2+2=4