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)
Answer:
[tex]f(x)=x^2-x+2[/tex]
Step-by-step explanation:
Quadratic equation form : [tex]y=ax^2+bx+c[/tex] --1
We are given points :(-1,4), (0,2) and (2,4).
Substitute the point (0,2) in the quadratic equation.
[tex]2=a(0)^2+b(0)+c[/tex]
[tex]2=c[/tex]
Thus the value of c is 2
Substitute the value of c in 1
Thus equation becomes: [tex]y=ax^2+bx+2[/tex] --2
Now substitute the point (-1,4) in 2
[tex]4=a(-1)^2+b(-1)+2[/tex]
[tex]4=a-b+2[/tex]
[tex]a-b=2[/tex] ---3
Now substitute point (2,4) in 2
[tex]4=a(2)^2+b(2)+2[/tex]
[tex]4=4a+2b+2[/tex]
[tex]2=2a+b+1[/tex]
[tex]2a+b=1[/tex] --4
Now solve 3 and 4 to find the value of a and b
Substitute the value of a from 3 in 4
[tex]2(2+b)+b=1[/tex]
[tex]4+2b+b=1[/tex]
[tex]4+3b=1[/tex]
[tex]3b=-3[/tex]
[tex]b=-1[/tex]
Substitute the value of b in 3
[tex]a-(-1)=2[/tex]
[tex]a+1=2[/tex]
[tex]a=2-1[/tex]
[tex]a=1[/tex]
Thus a = 1, b =-1 and c = 2
Substitute the values in 1
[tex]y=x^2-x+2[/tex]
Thus the quadratic function that contains the points (-1,4), (0,2) and (2,4) is [tex]f(x)=y=x^2-x+2[/tex]
Hence Option 3 is correct.
