Influence of Parameters in the Vertex Form

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

$f\left(x\right)=a(x-d{)}^2+ef\backslash left(x\backslash right)=a(x-d)^2+e$

from which you can read off the vertex $S(d\mathrm{\mid }e)S(d|e)$.

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

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.

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

$S(12\mathrm{\mid }-6)S(12|-6)$

Now, the vertex $S(d\mathrm{\mid }e)S(d|e)$ is contained in the vertex form $f(x)=a\cdot (x-d{)}^2+ef(x)=a\; \backslash cdot\; (x-d)^2+e$ .

$f(x)=a\cdot (x-12{)}^2-6f(x)=a\backslash cdot(x-12)^2-6$

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. $(6\mid 0)(6\mid 0)$ is a point on the graph of $ff$

$0=f(6)=a\cdot (6-12{)}^2-60=f(6)=a\backslash cdot(6-12)^2-6$

Solve this equation for $aa$.

$6=a(-6{)}^{2}6=a(-6)^2$

$a=\frac{1}{6}a=\backslash frac\{1\}\{6\}$

Substitute $aa$ into the equation of the function.

$f(x)=\frac{1}{6}(x-12{)}^2-6f(x)=\backslash frac\{1\}\{6\}(x-12)^2-6$

Visualisation in an applet

Use the sliders to change the parameters.

Video about the vertex form