Question

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

Answer 1

Answer:

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

Step-by-step 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.

Answer 2

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)