Astandard form of a parabola with points through (2, 0) (3, 2) (4, 6)

y = x^2 + 3x + 2
y = x^2 + 3x - 2
y = x^2 - 3x + 2
y = x^2 - 3x - 2

C. y = x^2 - 3x + 2

Step-by-step explanation:

(4,6)

6 = 4² - 3(4) + 2

6 = 16 - 12 + 2

6 = 6

2 = 2

(3,2)

2 = 3² - 3(3) + 2

2 = 9 - 9 + 2

2 = 2

(2,0)

0 = 2² - 3(2) + 2

0 = 4 - 6 + 2

0 = 0

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

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