第3章 3.2

(1)

(0,3)(0, -3)

y(3)=3x2y=3x23y - (-3) = -3x^2 \qquad y = -3x^2 - 3

(2)

(2,4)(2, 4)

y4=3(x2)2y=3(x2)2+4y - 4 = -3(x-2)^2 \qquad y = -3(x-2)^2 + 4

(3)

(2,4)(-2, 4)

y4=3(x(2))2y=3(x+2)2+4y - 4 = -3\left(x - (-2)\right)^2 \qquad y = -3(x+2)^2 + 4

(4)

(1,2)(-1, -2)

y(2)=3(x(1))2y=3(x+1)22y - (-2) = -3\left(x - (-1)\right)^2 \qquad y = -3(x+1)^2 - 2

解説: r31bn1z