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

Answer:

$y = 1 - x^2$

Step-by-step 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.