什么是几何中的顶点?
当两条或多条线在一点相遇形成一个角度时,该点可以称为顶点。所以顶点可以定义为两条或多条线相交形成一个角度的点。顶点用大写字母表示,如 A、B、E 等。在几何中有许多形状,如立方体、正方形、三角形等。对于这些图形,有多个顶点。所以顶点的复数形式称为顶点。让我们看几个数字
有一个欧拉公式可以计算一个三维 (3D) 图形存在多少个顶点。该公式由 -
Euler’s Formula-> F + V – E = 2
The above formula can be modified to get vertices count as
V = E + 2 – F
Where
V represents number of vertices
F represents number of faces
E represents number of edges
让我们看一下关于找到给定图形的顶点数的几个问题。
问题 1:求一个具有 6 个面和 12 个边的图形(立方体)存在的顶点数。
解决方案:
Given
Number of faces (F) = 6
Number of edges (E) = 12
From Euler’s Method,
Number of vertices (V) = E + 2 – F
= 12 + 2 – 6
= 8
So number of vertices for given figure is 8.
问题 2:查找具有 2 个面(覆盖顶部和底部)和 0 个边的 3D 圆柱体的顶点数。
解决方案:
Given
Number of faces (F) = 2
Number of edges (E) = 0
From Euler’s Method,
Number of vertices (V) = E + 2 – F
= 0 + 2 – 2
= 0
So number of vertices for given figure is 0.
抛物线的顶点
在抛物线中,顶点是它实际转动的点。这也称为最小点/最大点。当抛物线打开时,顶点称为最大值点,否则称为最小值点。
根据给定的方程形式,有两种方法可以在抛物线中找到顶点。
如果给定的抛物线方程的形式为 ax 2 +bx+c,则抛物线的顶点由下式给出 -
V = (-b/2a, -D/4a)
Where
D = b2 – 4ac
如果给定抛物线方程的形式为 y = a(x – h) 2 + k,则抛物线的顶点由下式给出 -
V = (h , k)
让我们看一些寻找抛物线顶点的例子。
问题 1:如果抛物线方程为 3x 2 + x – 2,求抛物线的顶点。
解决方案:
Given
Equation of parabola 3x2+x-2
It is of form ax2+bx+c where a=3, b=1, c=-2
So vertex of parabola is V=(-b/2a,-D/4a)
Discriminant can be calculate by formula D=b2-4ac
D=12-4×3×(-2)
=1-(-24)
=1+24
D=25
Vertex (V)=(-b/2a,-D/4a)
=(-1/2(3),-25/4(3))
V=(-1/6,-25/12)
Hence the vertex of parabola 3x2+x-2 is at (-1/6,-25/12)
问题2:如果抛物线方程是x 2 -4x+3,那么抛物线的顶点是多少。
解决方案:
Given
Equation of parabola is x2-4x+3
It is of form ax2+bx+c where a=1, b=-4, c=3
So vertex of parabola is V=(-b/2a,-D/4a)
Discriminant can be calculate by formula D=b2-4ac
D=(-4)2-4×1×3
=16-12
D=4
Vertex (V)=(-b/2a,-D/4a)
=(-(-4)/2(1),-4/4(1))
=(4/2,-4/4)
V=(2,-1)
Hence the vertex of parabola x2-4x+3 is at (2,-1)
问题3:如果抛物线方程为y = 3(x-4) 2 +2,求抛物线的顶点。
解决方案:
Given
Equation of parabola is y=3(x-4)2+2
It is of form y=a(x-h)2+k where a=3, h=4, k=2
So vertex of parabola is V=(h,k)
Vertex (V)=(h , k)
=(4,2)
Hence the vertex of parabola 3(x-4)2+2 is at (4,2)
问题4:如果抛物线方程为y = 2x 2 -8x+9,那么抛物线的顶点是多少。
解决方案:
Given
Equation of parabola is y=2x2-8x+9
This can be rewritten into y=2(x-2)2+1
It is of form y=a(x-h)2+k where a=2, h=2, k=1
So vertex of parabola is V=(h,k)
Vertex (V)=(h , k)
=(2,1)
Hence the vertex of parabola 2x2-8x+9 is at (2,1)