(1) y=3x2+px+q=3(x2+p3x)+q=3(x+p6)2−p212+qy = 3x^2 + px + q = 3\left(x^2 + \frac{p}{3}x\right) + q = 3\left(x + \frac{p}{6}\right)^2 - \frac{p^2}{12} + qy=3x2+px+q=3(x2+3px)+q=3(x+6p)2−12p2+q x=1 で最小値５となるので p6=−1p=−6\frac{p}{6} = -1 \qquad p = -66p=−1p=−6 −p212+q=−3+q=5q=8-\frac{p^2}{12} + q = -3 + q = 5 \qquad q = 8−12p2+q=−3+q=5q=8 これからp=−6p = -6p=−6、q=8q = 8q=8 (2) y=px2+x+q=p(x2+1px)+q=p(x+12p)2−14p+qy = px^2 + x + q = p\left(x^2 + \frac{1}{p}x\right) + q = p\left(x + \frac{1}{2p}\right)^2 - \frac{1}{4p} + qy=px2+x+q=p(x2+p1x)+q=p(x+2p1)2−4p1+q x=2x = 2x=2 で最大値１となるので 12p=−2p=−14\frac{1}{2p} = -2 \qquad p = -\frac{1}{4}2p1=−2p=−41 −14p+q=1+q=1q=0-\frac{1}{4p} + q = 1 + q = 1 \qquad q = 0−4p1+q=1+q=1q=0 これから p=−14p = -\dfrac{1}{4}p=−41、q=0q = 0q=0