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Influence of Parameters in the Vertex Form

Starting from the normal parabola, one can construct any parabola.

Starting from the normal parabola, one can construct any parabola. For this, you can use the vertex form:

f(x)=a(xd)2+ef\left(x\right)=a(x-d)^2+e

from which you can read off the vertex S(de)S(d|e).

The individual parameters a,d,ea,d,e of the vertex form have the following influence on the graph of the parabola:

Parameter

Influence on

Graph

a

Opening direction and extension/compression

  • a>0a>0: Parabola opened upwards

  • a<0a<0: Parabola opened downwards

  • a<1|a|<1: Parabola compressed in the y-axis direction (compared to the normal parabola)

  • a>1|a|>1: Parabola stretched in the y-axis direction (compared to the normal parabola)

Image

d

Shift in xx-direction

  • d>0d>0: Shift by dd to the right

  • d<0d<0: Shift by dd to the left

Image

e

Shift in yy-direction

  • e>0e>0: Shift by ee nupwards

  • e<0e<0: Shift by ee downwards

Image

Example

Finde zu der nebenstehenden Parabel in dem Koordinatensystem die zugehörige Funktionsgleichung, das heißt mit passenden Parametern a,d,ea,d,e. Betrachte in der obigen Tabelle nochmal, welche Auswirkungen die Parameter haben.

ParabelBsp

You can recognise the vertex of the parabola in the graph.

S(126)S(12|-6)

Now, the vertex S(de)S(d|e) is contained in the vertex form f(x)=a(xd)2+ef(x)=a \cdot (x-d)^2+e .

f(x)=a(x12)26f(x)=a\cdot(x-12)^2-6

The parameter aa remains to be determined. To do this, you can read off a point from the graph and insert it into the equation of the function. (60)(6∣0) is a point on the graph of ff

0=f(6)=a(612)260=f(6)=a\cdot(6-12)^2-6

Solve this equation for aa.

6=a(6)26=a(-6)^2

a=16a=\frac{1}{6}

Substitute aa into the equation of the function.

f(x)=16(x12)26f(x)=\frac{1}{6}(x-12)^2-6

Visualisation in an applet

Use the sliders to change the parameters.

Video about the vertex form


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