# Vertex

Vertices , short crest are in the geometry of special points on curves.

The vertices of a conic section ( ellipse , parabola or hyperbola ) are the points of intersection of the curve with the axes of symmetry. They are also the points at which the curvature is maximal or minimal.

The vertex of an upright parabola, which is a function graph of a quadratic function , is the high point or low point of the graph. The graph of a quadratic function is uniquely determined by the position of the vertex and the stretching factor. The computational determination of the vertex is therefore an important tool for drawing the graph of a quadratic function.

More generally, in differential geometry, a point on a regular curve is referred to as an apex or apex if the curvature there has a local extreme (i.e. a local maximum or minimum). The four vertex theorem makes a statement about the existence and the number of vertices in simply closed smooth plane curves.

## Vertex of a conic section

The vertices of a conic section are the intersections of such a curve with their axes of symmetry . The ellipse has four vertices, two main vertices and two secondary vertices, with the hyperbola there are two, with the parabola only one, the circle has no explicit vertex.

## Vertex of a parabola

The graph of a quadratic function is a parabola. Its apex is identical to the high point ( local maximum ) when it is open downwards and identical to the low point ( local minimum ) when it is open upwards.

If the position of the vertex is known, the parabola, insofar as it is a normal parabola , can be drawn quickly in a coordinate system with the aid of a parabola template . You can also use the parabola template to draw parabolas that are not normal parabolas if you scale the coordinate system accordingly.

### Vertex shape

Under the vertex shape or vertex shape of a quadratic function

${\ displaystyle f (x) \, = ax ^ {2} + bx + c {\ text {with}} a \ neq 0}$

one understands a certain form of this equation, from which one can read off the vertex of the function directly.

it is

${\ displaystyle f (x) = a (xd) ^ {2} + e}$

with the vertex . ${\ displaystyle S (d | e)}$

Consequently, the function in the form ${\ displaystyle f (x) = ax ^ {2} + bx + c {\ text {with}} a \ neq 0}$

${\ displaystyle f (x) = a \ left (x + {\ frac {b} {2a}} \ right) ^ {2} + {\ frac {4ac-b ^ {2}} {4a}}}$

be convicted.

The vertex then reads

${\ displaystyle S \ left (- {\ frac {b} {2a}} \ {\ Bigg |} \ c - {\ frac {b ^ {2}} {4a}} \ right).}$

In school, this formula is mostly not taught due to its size. Instead, the quadratic complement is taught, with the help of which one converts a quadratic function from the polynomial form to the vertex form.

#### Derivation by means of displacement

The normal parabola has its vertex in the coordinate origin. A stretching in the y-direction with the stretching factor (parabolic equation ) does not change anything. If this parabola is now shifted by units in the x direction and by units in the y direction so that its vertex has the coordinates , this can be represented using the following transformation: ${\ displaystyle y = x ^ {2}}$${\ displaystyle a}$${\ displaystyle y = ax ^ {2}}$${\ displaystyle d}$${\ displaystyle e}$${\ displaystyle S (d, e)}$

${\ displaystyle (ye) = a (xd) ^ {2}}$.

By multiplying we get:

${\ displaystyle ye = a (x ^ {2} -2dx + d ^ {2}) = ax ^ {2} -2adx + ad ^ {2}}$and from it .${\ displaystyle y = ax ^ {2} -2adx + ad ^ {2} + e}$

Comparison with the standard function equation yields: ${\ displaystyle f (x) = ax ^ {2} + bx + c}$

${\ displaystyle b = -2ad}$and .${\ displaystyle c = ad ^ {2} + e}$

This can be transformed too

${\ displaystyle d = {\ frac {b} {- 2a}}}$or .${\ displaystyle e = c-ad ^ {2} = ca \ left ({\ frac {b} {- 2a}} \ right) ^ {2} = c - {\ frac {ab ^ {2}} {4a ^ {2}}} = c - {\ frac {b ^ {2}} {4a}}}$

#### Derivation using a square extension

The above formula can be derived using the square complement. The general shape is reshaped to the vertex shape.

{\ displaystyle {\ begin {aligned} f (x) & = ax ^ {2} + bx + c \\ & = a \ left (x ^ {2} + {\ frac {b} {a}} x \ right) + c \\ & = a \ left (\ underbrace {x ^ {2} +2 \, {\ frac {b} {2a}} x + \ left ({\ frac {b} {2a}} \ right ) ^ {2}} - \ left ({\ frac {b} {2a}} \ right) ^ {2} \ right) + c \\ & = a \ left (\ qquad \ left (x + {\ frac { b} {2a}} \ right) ^ {2} \ quad - \ quad \ left ({\ frac {b} {2a}} \ right) ^ {2} \ right) + c \\ & = a \ left (x + {\ frac {b} {2a}} \ right) ^ {2} -a {\ frac {b ^ {2}} {4a ^ {2}}} + c \\ & = a \ left (x + {\ frac {b} {2a}} \ right) ^ {2} + {\ frac {4ac-b ^ {2}} {4a}} \ end {aligned}}}

From the coordinates of the vertex can be read directly: . ${\ displaystyle d = - {\ frac {b} {2a}}, \ quad e = {\ frac {4ac-b ^ {2}} {4a}} = c - {\ frac {b ^ {2}} {4a}}}$

#### Derivation by means of derivation

Since the slope at the vertex is equal to 0, it is possible to derive the above formula using the first derivative .

{\ displaystyle {\ begin {aligned} f (x) & = ax ^ {2} + bx + c \\ f '(x) & = 2ax + b \\ f' (d) & = 0 \\ 0 & = 2ad + b \ Rightarrow d = - {\ frac {b} {2a}} \ end {aligned}}}

Inserting in the normal form:

{\ displaystyle {\ begin {aligned} e & = ad ^ {2} + bd + c \\ & = a \ left (- {\ frac {b} {2a}} \ right) ^ {2} + b \ left (- {\ frac {b} {2a}} \ right) + c \\ & = {\ frac {b ^ {2}} {4a}} - {\ frac {2b ^ {2}} {4a}} + c \\ & = {\ frac {b ^ {2} -2b ^ {2}} {4a}} + c \\ & = c - {\ frac {b ^ {2}} {4a}} \ end {aligned}}}

### Examples

Diagram for example 1

example 1

${\ displaystyle f (x) = x ^ {2} -6x + 4}$

has the apex

${\ displaystyle S \ left (- {\ frac {-6} {2}} \ {\ Bigg |} {\ frac {4 \ cdot 4 - (- 6) ^ {2}} {4}} \ right) }$, so ${\ displaystyle S (3 | -5).}$

Example 2

${\ displaystyle f (x) = - x ^ {2} + 3x + 4}$

With , and the vertex is calculated as ${\ displaystyle a = -1}$${\ displaystyle b = 3}$${\ displaystyle c = 4}$

${\ displaystyle S \ left (- {\ frac {3} {2 \ cdot (-1)}} \ {\ Bigg |} \ {\ frac {4 \ cdot (-1) \ cdot 4-3 ^ {2 }} {4 \ cdot (-1)}} \ right)}$, so ${\ displaystyle S \ left ({\ frac {3} {2}} \ {\ Bigg |} \ {\ frac {25} {4}} \ right).}$

### Determination of the zeros from the vertex shape

The zeros of the respective quadratic function can be determined very easily from the vertex shape.

If you substitute with and with , you get the shape with the vertex . ${\ displaystyle - {\ frac {b} {2a}}}$${\ displaystyle d}$${\ displaystyle {\ frac {4ac-b ^ {2}} {4a}}}$${\ displaystyle e}$${\ displaystyle f (x) = a (xd) ^ {2} + e}$${\ displaystyle S (d | e)}$

Determination of the zeros:

{\ displaystyle {\ begin {aligned} f (x) & = 0 \\ [1ex] a (xd) ^ {2} + e & = 0 \\ [1ex] a (xd) ^ {2} & = - e \\ [1ex] (xd) ^ {2} & = - {\ frac {e} {a}} \\ [1ex] xd & = \ pm {\ sqrt {- {\ frac {e} {a}}} } \\ [1ex] x & = d \ pm {\ sqrt {- {\ frac {e} {a}}}} \ end {aligned}}}

Replacing and again with and results in the abc formula : ${\ displaystyle d}$${\ displaystyle e}$${\ displaystyle - {\ frac {b} {2a}}}$${\ displaystyle {\ frac {4ac-b ^ {2}} {4a}}}$

${\ displaystyle x = - {\ frac {b} {2a}} \ pm {\ sqrt {- {\ frac {\ frac {4ac-b ^ {2}} {4a}} {a}}}} = - {\ frac {b} {2a}} \ pm {\ sqrt {\ frac {b ^ {2} -4ac} {4a ^ {2}}}} = - {\ frac {b} {2a}} \ pm {\ frac {\ sqrt {b ^ {2} -4ac}} {2a}} = {\ frac {-b \ pm {\ sqrt {b ^ {2} -4ac}}} {2a}}}$