Lineare Funktion aus Punkten konstruieren
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
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Exercise:
Die Punkte A- und B- liegen auf dem Graphen der linearen Funktion fx. abclist abc Zeichne die beiden Punkte im folgen Koordinatensystem ein und beschrifte sie. abc Skizziere den Funktionsgraphen von fx im folgen Koordinatensystem. abc Bestimme die Funktionsgleichung analytisch. abc Bestimme die Umkehrfunktion von fx. Die Umkehrfunktion muss nicht skizziert werden! abclist center tikzpicture tkzInitxmin- xmax ymin- ymax xstep ystep tkzGridsub subxstep subystep tkzDrawXright tkzDrawYabove labelyfx tkzLabelXY tikzpicture center
Solution:
abclist abc siehe Koordinatensystem abc siehe Koordinatensystem abc Die Steigung der Funktion beträgt al m fracDelta yDelta x frac---- frac- -. Der Ordinatenabschnitt ist folglich al f - + q - + q - q . Die Funktionsgleichung ist damit al fx -x + . abc Wir lösen die Funktionsgleichung nach x auf: al y -x + uf +x -y x -y + uf : x -fracy+frac f^-x -fracx+frac abclist center tikzpicture tkzInitxmin- xmax ymin- ymax xstep ystep tkzGridsub subxstep subystep tkzDrawXright tkzDrawYabove labelyfx tkzLabelXY tkzFctdarkred very thick domain-:-*x+ tkzDefPoByFct- tkzDrawPossizetkzPoResult tkzLabelPobelow lefttkzPoResultA tkzDefPoByFct tkzDrawPossizetkzPoResult tkzLabelPoabove righttkzPoResultB tikzpicture center
Die Punkte A- und B- liegen auf dem Graphen der linearen Funktion fx. abclist abc Zeichne die beiden Punkte im folgen Koordinatensystem ein und beschrifte sie. abc Skizziere den Funktionsgraphen von fx im folgen Koordinatensystem. abc Bestimme die Funktionsgleichung analytisch. abc Bestimme die Umkehrfunktion von fx. Die Umkehrfunktion muss nicht skizziert werden! abclist center tikzpicture tkzInitxmin- xmax ymin- ymax xstep ystep tkzGridsub subxstep subystep tkzDrawXright tkzDrawYabove labelyfx tkzLabelXY tikzpicture center
Solution:
abclist abc siehe Koordinatensystem abc siehe Koordinatensystem abc Die Steigung der Funktion beträgt al m fracDelta yDelta x frac---- frac- -. Der Ordinatenabschnitt ist folglich al f - + q - + q - q . Die Funktionsgleichung ist damit al fx -x + . abc Wir lösen die Funktionsgleichung nach x auf: al y -x + uf +x -y x -y + uf : x -fracy+frac f^-x -fracx+frac abclist center tikzpicture tkzInitxmin- xmax ymin- ymax xstep ystep tkzGridsub subxstep subystep tkzDrawXright tkzDrawYabove labelyfx tkzLabelXY tkzFctdarkred very thick domain-:-*x+ tkzDefPoByFct- tkzDrawPossizetkzPoResult tkzLabelPobelow lefttkzPoResultA tkzDefPoByFct tkzDrawPossizetkzPoResult tkzLabelPoabove righttkzPoResultB tikzpicture center
Meta Information
Exercise:
Die Punkte A- und B- liegen auf dem Graphen der linearen Funktion fx. abclist abc Zeichne die beiden Punkte im folgen Koordinatensystem ein und beschrifte sie. abc Skizziere den Funktionsgraphen von fx im folgen Koordinatensystem. abc Bestimme die Funktionsgleichung analytisch. abc Bestimme die Umkehrfunktion von fx. Die Umkehrfunktion muss nicht skizziert werden! abclist center tikzpicture tkzInitxmin- xmax ymin- ymax xstep ystep tkzGridsub subxstep subystep tkzDrawXright tkzDrawYabove labelyfx tkzLabelXY tikzpicture center
Solution:
abclist abc siehe Koordinatensystem abc siehe Koordinatensystem abc Die Steigung der Funktion beträgt al m fracDelta yDelta x frac---- frac- -. Der Ordinatenabschnitt ist folglich al f - + q - + q - q . Die Funktionsgleichung ist damit al fx -x + . abc Wir lösen die Funktionsgleichung nach x auf: al y -x + uf +x -y x -y + uf : x -fracy+frac f^-x -fracx+frac abclist center tikzpicture tkzInitxmin- xmax ymin- ymax xstep ystep tkzGridsub subxstep subystep tkzDrawXright tkzDrawYabove labelyfx tkzLabelXY tkzFctdarkred very thick domain-:-*x+ tkzDefPoByFct- tkzDrawPossizetkzPoResult tkzLabelPobelow lefttkzPoResultA tkzDefPoByFct tkzDrawPossizetkzPoResult tkzLabelPoabove righttkzPoResultB tikzpicture center
Die Punkte A- und B- liegen auf dem Graphen der linearen Funktion fx. abclist abc Zeichne die beiden Punkte im folgen Koordinatensystem ein und beschrifte sie. abc Skizziere den Funktionsgraphen von fx im folgen Koordinatensystem. abc Bestimme die Funktionsgleichung analytisch. abc Bestimme die Umkehrfunktion von fx. Die Umkehrfunktion muss nicht skizziert werden! abclist center tikzpicture tkzInitxmin- xmax ymin- ymax xstep ystep tkzGridsub subxstep subystep tkzDrawXright tkzDrawYabove labelyfx tkzLabelXY tikzpicture center
Solution:
abclist abc siehe Koordinatensystem abc siehe Koordinatensystem abc Die Steigung der Funktion beträgt al m fracDelta yDelta x frac---- frac- -. Der Ordinatenabschnitt ist folglich al f - + q - + q - q . Die Funktionsgleichung ist damit al fx -x + . abc Wir lösen die Funktionsgleichung nach x auf: al y -x + uf +x -y x -y + uf : x -fracy+frac f^-x -fracx+frac abclist center tikzpicture tkzInitxmin- xmax ymin- ymax xstep ystep tkzGridsub subxstep subystep tkzDrawXright tkzDrawYabove labelyfx tkzLabelXY tkzFctdarkred very thick domain-:-*x+ tkzDefPoByFct- tkzDrawPossizetkzPoResult tkzLabelPobelow lefttkzPoResultA tkzDefPoByFct tkzDrawPossizetkzPoResult tkzLabelPoabove righttkzPoResultB tikzpicture center
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