y=ax2+3x−4=a(x2+3ax)−4=a(x+32a)2−94a−4y = ax^2 + 3x - 4 = a\left(x^2 + \dfrac{3}{a}x\right) - 4 = a\left(x + \dfrac{3}{2a}\right)^2 - \dfrac{9}{4a} - 4y=ax2+3x−4=a(x2+a3x)−4=a(x+2a3)2−4a9−4 y=2x2−6x+b=2(x2−3x)+b=2(x−32)2−92+by = 2x^2 - 6x + b = 2(x^2 - 3x) + b = 2\left(x - \dfrac{3}{2}\right)^2 - \dfrac{9}{2} + by=2x2−6x+b=2(x2−3x)+b=2(x−23)2−29+b 頂点が一致するので 32a=−32a=−1\dfrac{3}{2a} = -\dfrac{3}{2} \qquad a = -12a3=−23a=−1 −94a−4=−92+b94−4=−92+bb=114-\dfrac{9}{4a} - 4 = -\dfrac{9}{2} + b \qquad \dfrac{9}{4} - 4 = -\dfrac{9}{2} + b \qquad b = \dfrac{11}{4}−4a9−4=−29+b49−4=−29+bb=411 これから a=−1、 b=114a = -1、\ b = \dfrac{11}{4}a=−1、 b=411