y=ax^2 + bx + c
=a(x^2 + (b/a) *x +c/a)
=a(x^2 + (b/a)*x + (b/2a)^2 - (b/2a)^2 + c/a)
=a[(x+b/2a)^2 + (4ac-b^2)/4a^2]
=a(x+b/2a)^2 + (4ac-b^2)/4a
1、這三個式子都是一樣的,是換了形式的一元二次函式的表示式(Quadratic Expression).
2、通式是 y = ax^2 + bx + c
下面幫你推導一下其它的兩個
y = ax^2 + bx + c
= a[x^2 + (b/a)x] + c
= a(x + b/2a)^2 - a*(b/2a)^2 + c
= a(x + b/2a)^2 -b^2/4a + c
= a(x + b/2a)^2 -(b^2 - 4ac)/4a
= a(x + b/2a)^2 + (4ac - b^2)/4a
說明:a.這種方法叫“配方法”(Completing Square)
b.這種方法用來解方程,或找到曲線與x軸的交點。
應用上面的結果:
y = a(x + b/2a)^2 -(b^2 - 4ac)/4a
b^2 - 4ac ≥ 0
b^2 - 4ac = [根號下(b^2 - 4ac)]^2
∴ y = a(x + b/2a)^2 -(b^2 - 4ac)/4a
= a{(x + b/2a)^2 -[根號下(b^2 - 4ac)]^2 /4a^2}
= a{x+b/2a+[根號下(b^2-4ac)]/2a}{x+b/2a-[根號下(b^2-4ac)]/2a}
= a{x-[-b+根號下(b^2-4ac)]/2a}{x+[-b+根號下(b^2-4ac)]/2a}
= a{x - x1}{x - x2}
這裡的x1=[-b+根號下(b^2-4ac)]/2a, x2=[-b-根號下(b^2-4ac)]/2a
y=ax^2 + bx + c
=a(x^2 + (b/a) *x +c/a)
=a(x^2 + (b/a)*x + (b/2a)^2 - (b/2a)^2 + c/a)
=a[(x+b/2a)^2 + (4ac-b^2)/4a^2]
=a(x+b/2a)^2 + (4ac-b^2)/4a
1、這三個式子都是一樣的,是換了形式的一元二次函式的表示式(Quadratic Expression).
2、通式是 y = ax^2 + bx + c
下面幫你推導一下其它的兩個
y = ax^2 + bx + c
= a[x^2 + (b/a)x] + c
= a(x + b/2a)^2 - a*(b/2a)^2 + c
= a(x + b/2a)^2 -b^2/4a + c
= a(x + b/2a)^2 -(b^2 - 4ac)/4a
= a(x + b/2a)^2 + (4ac - b^2)/4a
說明:a.這種方法叫“配方法”(Completing Square)
b.這種方法用來解方程,或找到曲線與x軸的交點。
應用上面的結果:
y = a(x + b/2a)^2 -(b^2 - 4ac)/4a
b^2 - 4ac ≥ 0
b^2 - 4ac = [根號下(b^2 - 4ac)]^2
∴ y = a(x + b/2a)^2 -(b^2 - 4ac)/4a
= a{(x + b/2a)^2 -[根號下(b^2 - 4ac)]^2 /4a^2}
= a{x+b/2a+[根號下(b^2-4ac)]/2a}{x+b/2a-[根號下(b^2-4ac)]/2a}
= a{x-[-b+根號下(b^2-4ac)]/2a}{x+[-b+根號下(b^2-4ac)]/2a}
= a{x - x1}{x - x2}
這裡的x1=[-b+根號下(b^2-4ac)]/2a, x2=[-b-根號下(b^2-4ac)]/2a