Answer:
[tex]y = 1 - x^2[/tex]
Step-by-step explanation:
we're given three points (0,1),(1,0) and (-1,0). and and equation of a parabola
[tex]y = a + bx + cx^2[/tex]
we can plug in each of the coordinates, and 3 equations
(x,y) = (0,1)
[tex]1 = a + b(0) + c(0)^2[/tex]
[tex]a=1\quad\quad\Rightarrow A[/tex]
(x,y) = (1,0)
[tex]0 = a + b(1) + c(1)^2[/tex]
[tex]a + b + c=0\quad\quad\Rightarrow B[/tex]
(x,y) = (-1,0)
[tex]0 = a + b(-1) + c(-1)^2[/tex]
[tex]a - b + c=0\quad\quad\Rightarrow C[/tex]
These are are three equations, well we can simultaneously solve them to find the values of a, b and c.
we already found that, a = 1. so we plug this value in the rest of the equations. we'll use equation B.
[tex]a + b + c=0[/tex]
[tex]1 + b + c=0[/tex]
[tex]b=-1-c[/tex]
we can substitute this value of b and a = 1, equation C
[tex]a - b + c=0[/tex]
[tex]1 -(-1-c) + c=0[/tex]
[tex]1 +1+c + c=0[/tex]
[tex]2+2c=0[/tex]
[tex]c=-1[/tex]
we can use this value of c back in b
[tex]b=-1-(-1)[/tex]
[tex]b=0[/tex]
hence our equation of the 2-degree polynomial will be:
[tex]y = a + bx + cx^2[/tex]
[tex]y = 1 + (0)x + (-1)x^2[/tex]
[tex]y = 1 - x^2[/tex]
and this polynomial indeed passes through all the points (0, 1), (1, 0), and (-1, 0).
And since these are the only solutions to the simultaneous equation we solved (i.e. we have single values of a,b and c). there's no other possibility.
