本文搬运自本人初中博客园博客，若图片加载不出来，可到原文查看：https://www.cnblogs.com/zhangtingxi/p/16473421.html

y=ax2+bx+c\Large y=ax^2+bx+c

y=a(x2+bax+ca)\Large y=a(x^2+\dfrac{b}{a}x+\dfrac{c}{a})

y=a(x2+2×x×b2a+ca)\Large y=a(x^2+2\times x\times \dfrac b {2a}+\dfrac c a)

y=a[x2+2xb2a+(b2a)2−(b2a)2+ca]\Large y=a[x^2+2x\dfrac b {2a}+(\dfrac b {2a})^2-(\dfrac b {2a})^2+\dfrac c a]

y=a[(x+b2a)2−b24a2+ca]\Large y=a[(x+\dfrac b {2a})^2-\dfrac{b^2}{4a^2}+\dfrac c a]

y=a(x+b2a)2−b24a+c\Large y=a(x+\dfrac b {2a})^2-\dfrac{b^2}{4a}+c

y=a(x+b2a)2−b24a+4ac4a\Large y=a(x+\dfrac b {2a})^2-\dfrac{b^2}{4a}+\dfrac {4ac}{4a}

y=a(x+b2a)2+4ac−b24a\Large y=a(x+\dfrac b {2a})^2+\dfrac{4ac-b^2}{4a}

因此其顶点坐标为 (−b2a,4ac−b24a)\Large (-\dfrac b {2a},\dfrac{4ac-b^2}{4a})