中考数学已知抛物线Y=1/2X2+X-
配方法:
ax^+bx+c=a[x^+(b/a)x+c/a]
=a[x^+2×(b/2a)×x+c/a]
=a[x^+2×(b/2a)×x+(b/2a)^-(b/2a)^+c/a]
=a[(x+b/2a)^+4ca/4a^-b^/4a^]
=a[(x+b/2a)^+(4ac-b^)/4a^]
=a×(x+b/2a)^+(4ac-b^)/4a
顶点坐标x=-b/2a y=(4ac-b^)/4a
ax^+bx+c=a×(x+b/2a)^+(4ac-b^)/4a=0
(x+b/2a)^=(b^-4ac)/a^
x1=[√(b^-4ac)]/a-b/2a=[-b+√(b^-4ac)]/2a
x...全部
配方法:
ax^+bx+c=a[x^+(b/a)x+c/a]
=a[x^+2×(b/2a)×x+c/a]
=a[x^+2×(b/2a)×x+(b/2a)^-(b/2a)^+c/a]
=a[(x+b/2a)^+4ca/4a^-b^/4a^]
=a[(x+b/2a)^+(4ac-b^)/4a^]
=a×(x+b/2a)^+(4ac-b^)/4a
顶点坐标x=-b/2a y=(4ac-b^)/4a
ax^+bx+c=a×(x+b/2a)^+(4ac-b^)/4a=0
(x+b/2a)^=(b^-4ac)/a^
x1=[√(b^-4ac)]/a-b/2a=[-b+√(b^-4ac)]/2a
x2=-[√(b^-4ac)]/a-b/2a=[-b-√(b^-4ac)]/2a
x1,x2就是求根公式。
x1,x2是抛物线与X轴交点的坐标。
│AB│=│x1-x2│=[√(b^-4ac)]/2│a│
带入即可了。
。收起