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?

Step-by-step explanation:

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

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

(x,y) = (0,1)

(x,y) = (1,0)

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

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.

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

we can use this value of c back in b

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

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.

need points to ask question

step-by-step explanation:

trapezoid

step-by-step explanation:

first i plotted all my points on a grid graph

i made sure all of them were in the correct quadrant

then i saw it was a trapezoid or a quadrilateral