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Formulate the quadratic function that contains the points (-1,4), (0,2) and (2,4).

f(x) = x2 + x + 2
f(x) = x2 + x - 2
f(x) = x2 - x + 2
f(x) = x2 - x - 2

2 Answers

4 votes

Answer:


f(x)=x^2-x+2

Explanation:

Quadratic equation form :
y=ax^2+bx+c --1

We are given points :(-1,4), (0,2) and (2,4).

Substitute the point (0,2) in the quadratic equation.


2=a(0)^2+b(0)+c


2=c

Thus the value of c is 2

Substitute the value of c in 1

Thus equation becomes:
y=ax^2+bx+2 --2

Now substitute the point (-1,4) in 2


4=a(-1)^2+b(-1)+2


4=a-b+2


a-b=2 ---3

Now substitute point (2,4) in 2


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


4=4a+2b+2


2=2a+b+1


2a+b=1 --4

Now solve 3 and 4 to find the value of a and b

Substitute the value of a from 3 in 4


2(2+b)+b=1


4+2b+b=1


4+3b=1


3b=-3


b=-1

Substitute the value of b in 3


a-(-1)=2


a+1=2


a=2-1


a=1

Thus a = 1, b =-1 and c = 2

Substitute the values in 1


y=x^2-x+2

Thus the quadratic function that contains the points (-1,4), (0,2) and (2,4) is
f(x)=y=x^2-x+2

Hence Option 3 is correct.

User Unutbu
by
8.4k points
4 votes

f(x) = x2 - x + 2

f(-1)=(-1)²-(-1)+2=4 contain the point (-1,4)

f(0) = 0²-0+2 = 2 contain the point (0,2)

f(2) = 2²-2+2 =4 contain the point (2,4)

User Paul Keeble
by
7.5k points

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