Question

Find a second-degree polynomial (that is, an equation of the form y =a + bx + cx2 d that goes through the three points (0, 1), (1, 0), and (-1, 0). Is there more than one possibility?

Answer 1

`y = 1 - x^2`

Explanation:

we're given three points (0,1),(1,0) and (-1,0). and and equation of a parabola

`y = a + bx + cx^2`

we can plug in each of the coordinates, and 3 equations

(x,y) = (0,1)

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

`a=1\quad\quad\Rightarrow A`

(x,y) = (1,0)

`0 = a + b(1) + c(1)^2`

`a + b + c=0\quad\quad\Rightarrow B`

(x,y) = (-1,0)

`0 = a + b(-1) + c(-1)^2`

`a - b + c=0\quad\quad\Rightarrow C`

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.

`a + b + c=0`

`1 + b + c=0`

`b=-1-c`

we can substitute this value of b and a = 1, equation C

`a - b + c=0`

`1 -(-1-c) + c=0`

`1 +1+c + c=0`

`2+2c=0`

`c=-1`

we can use this value of c back in b

`b=-1-(-1)`

`b=0`

hence our equation of the 2-degree polynomial will be:

`y = a + bx + cx^2`

`y = 1 + (0)x + (-1)x^2`

`y = 1 - x^2`

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.