Content

• Steps
• Method 1 of 5: Find the number of vertices in a polyhedron
• Method 2 of 5: Finding the vertex of the domain of a system of linear inequalities
• Method 3 of 5: Finding the vertex of a parabola through the axis of symmetry
• Method 4 of 5: Finding the vertex of a parabola by completing a square
• Method 5 of 5: Find the vertex of a parabola using a simple formula
• What do you need

In mathematics, there are a number of problems in which you need to find the top. For example, a vertex of a polyhedron, a vertex or several vertices of a domain of a system of inequalities, a vertex of a parabola or quadratic equation. This article will show you how to find the top in different problems.

Steps

Method 1 of 5: Find the number of vertices in a polyhedron

1. 1 Euler's theorem. The theorem states that in any polyhedron, the number of its vertices plus the number of its faces minus the number of its edges is always two.
   • Formula describing Euler's theorem: F + V - E = 2
     • F is the number of faces.
     • V is the number of vertices.
     • E is the number of ribs.
2. 2 Rewrite the formula to find the number of vertices. Given the number of faces and the number of edges of a polyhedron, you can quickly find the number of vertices using Euler's formula.
   • V = 2 - F + E
3. 3 Plug the values ​​you give into this formula. This gives you the number of vertices in the polyhedron.
   • Example: Find the number of vertices of a polyhedron that has 6 faces and 12 edges.
     • V = 2 - F + E
     • V = 2 - 6 + 12
     • V = -4 + 12
     • V = 8

Method 2 of 5: Finding the vertex of the domain of a system of linear inequalities

1. 1 Plot the solution (area) of a system of linear inequalities. In certain cases, you can see some or all of the vertices of the area of ​​the system of linear inequalities on the graph. Otherwise, you have to find the vertex algebraically.
   • When using a graphing calculator, you can view the entire graph and find the coordinates of the vertices.
2. 2 Convert inequalities to equations. In order to solve the system of inequalities (that is, find "x" and "y"), you need to put an "equal" sign instead of the inequality signs.
   • Example: given a system of inequalities:
     • y x
     • y> - x + 4
   • Convert inequalities to equations:
     • y = x
     • y = - x + 4
3. 3 Now express any variable in one equation and plug it into another equation. In our example, plug the y value from the first equation into the second equation.
   • Example:
     • y = x
     • y = - x + 4
   • Substitute y = x in y = - x + 4:
     • x = - x + 4
4. 4 Find one of the variables. Now you have an equation with only one variable, x, which is easy to find.
   • Example: x = - x + 4
     • x + x = 4
     • 2x = 4
     • 2x / 2 = 4/2
     • x = 2
5. 5 Find another variable. Substitute the found value "x" in any of the equations and find the value "y".
   • Example: y = x
     • y = 2
6. 6 Find the top. The vertex has coordinates equal to the found values ​​"x" and "y".
   • Example: the vertex of the region of the given system of inequalities is the point O (2,2).

Method 3 of 5: Finding the vertex of a parabola through the axis of symmetry

1. 1 Factor the equation. There are several ways to factor a quadratic equation. As a result of the expansion, you get two binomials, which, when multiplied, will lead to the original equation.
   • Example: given a quadratic equation
     • 3x2 - 6x - 45
     • First, bracket the common factor: 3 (x2 - 2x - 15)
     • Multiply the coefficients "a" and "c": 1 * (-15) = -15.
     • Find two numbers, the multiplication of which is -15, and their sum is equal to the coefficient "b" (b = -2): 3 * (-5) = -15; 3 - 5 = -2.
     • Plug the found values ​​into the equation ax2 + kx + hx + c: 3 (x2 + 3x - 5x - 15).
     • Expand the original equation: f (x) = 3 * (x + 3) * (x - 5)
2. 2 Find the point (s) at which the graph of the function (in this case, the parabola) crosses the abscissa. The graph crosses the X-axis at f (x) = 0.
   • Example: 3 * (x + 3) * (x - 5) = 0
     • x +3 = 0
     • x - 5 = 0
     • x = -3; x = 5
     • Thus, the roots of the equation (or points of intersection with the X-axis): A (-3, 0) and B (5, 0)
3. 3 Find the axis of symmetry. The axis of symmetry of the function passes through a point that lies in the middle between the two roots. In this case, the vertex lies on the axis of symmetry.
   • Example: x = 1; this value lies in the middle between -3 and +5.
4. 4 Plug in the x value into the original equation and find the y value. These "x" and "y" values ​​are the coordinates of the vertex of the parabola.
   • Example: y = 3x2 - 6x - 45 = 3 (1) 2 - 6 (1) - 45 = -48
5. 5 Write down your answer.
   • Example: the vertex of a given quadratic equation is the point O (1, -48)

Method 4 of 5: Finding the vertex of a parabola by completing a square

1. 1 Rewrite the original equation as: y = a (x - h) ^ 2 + k, while the vertex lies at the point with coordinates (h, k). To do this, you need to supplement the original quadratic equation to a complete square.
   • Example: given a quadratic function y = - x ^ 2 - 8x - 15.
2. 2 Consider the first two terms. Factor out the coefficient of the first term (the intercept is ignored).
   • Example: -1 (x ^ 2 + 8x) - 15.
3. 3 Expand the free term (-15) into two numbers so that one of them completes the expression in parentheses to a complete square. One of the numbers must be equal to the square of half the coefficient of the second term (from the expression in parentheses).
   • Example: 8/2 = 4; 4 * 4 = 16; so
     • -1 (x ^ 2 + 8x + 16)
     • -15 = -16 + 1
     • y = -1 (x ^ 2 + 8x + 16) + 1
4. 4 Simplify the equation. Since the expression in brackets is a complete square, you can rewrite this equation in the following form (if necessary, perform addition or subtraction operations outside the brackets):
   • Example: y = -1 (x + 4) ^ 2 + 1
5. 5 Find the coordinates of the vertex. Recall that the coordinates of the vertex of a function of the form y = a (x - h) ^ 2 + k are (h, k).
   • k = 1
   • h = -4
   • Thus, the vertex of the original function is the point O (-4,1).

Method 5 of 5: Find the vertex of a parabola using a simple formula

1. 1 Find the "x" coordinate using the formula: x = -b / 2a (for a function of the form y = ax ^ 2 + bx + c). Plug in the "a" and "b" values ​​into the formula and find the "x" coordinate.
   • Example: given a quadratic function y = - x ^ 2 - 8x - 15.
   • x = -b / 2a = - (- 8) / (2 * (- 1)) = 8 / (- 2) = -4
   • x = -4
2. 2 Plug in the x value you find into the original equation. Thus, you will find "y". These "x" and "y" values ​​are the coordinates of the vertex of the parabola.
   • Example: y = - x ^ 2 - 8x - 15 = - (- 4) ^ 2 - 8 (-4) - 15 = - (16) - (- 32) - 15 = -16 + 32 - 15 = 1
     • y = 1
3. 3 Write down your answer.
   • Example: the vertex of the original function is the point O (-4,1).

What do you need

• Calculator
• Pencil
• Paper