Kann mir einer Schritt für Schritt erklären, wie man es umwandelt . Wenn ich im Internet danach suche, kriege nur Beispiele, wo vor dem x eine zahl steht, die ich ausklammern kann und halbieren muss. Ich weiß nicht, wie ich x halbieren soll oder x^2 ausklammern kann
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Loading...Der Graph einer ganz rationalen Funktion 3. Grades schneidet die Y-Achse bei -1 die Tangente in diesem Schnittpunkt ist die Y-Achse. Der Graph hat zusätzlich den Tiefpunkt 0 / -1 Ich weis das ich 4 Bedingungen habe aber ich habe nur 3 Punkte ?
Bei der theoretischen Wahrscheinlichkeit rechnet man ja die Anzahl der günstigen Ergebnisse eines Ereignisses durch die Anzahl aller möglichen Ergebnisse. Aber das heißt doch noch lange nicht, dass alle Ergebnisse gleichwahrscheinlich sind, wie bei der Laplace Wahrscheinlichkeit,oder??
Wie berechnet man diese Aufgabe? Zinseszinsen Ko= 1200€ p%=3,9% n= 8 Jahre Kn= ?
Hallo, Ich bin mir Sicher das Experiment 1 zur Verteilung (2) gehört. Ich bin mir auch sicher, dass Experiment 2 zu Verteilung (3) gehört. ABER Bei Experiment 3 komme ich nicht zur Veteilung wie bei (1)
Hallo wie groß ist Winkel Beta? Aufgabe 12. Und wie soll ich die Dreiecke im Aufgabe 10 beschriften?
Hello,
the general form of a parabola is ax2+bx+c.
The vertex is where the derivation is equal to zero.
Discharge is generally f'(x)=2ax+b.
2ax+b=0; 2ax=-b; x=-b/(2a).
For x=-b/(2a), therefore, the vertex of a parabola lies.
You get the y coordinate of the vertex when you use -b/(2a) instead of x in f(x):
f(-b/(2a))=a*(-b/(2a))2+b*(-b/(2a)))+c.
This can be summarized to a*b2/4a2)-b2/(2a)+c and further to
b2/(4a)-b2/(2a)+c.
Extend -b2/(2a) with 2, you can combine it with b2/(4a) to -b2/(4a).
If you expand c with 4a, you can put everything on a break:
(4ac-b2)/(4a).
This is the y coordinate of the vertex of an x-popular parabola.
The vertex is therefore S (-b/(2a)|(4ac-b2)/(4a).
Now you can convert the general form over the square supplement.
First you draw the factor a from ax2+bx+c:
a*(x2+(b/a)x+c/a).
Now you convert the first two summands in the clamp by the square addition (b/(2a))2=b2/(4a2) to x2+(b/a)x+b2/(4a2)=(x+b/(2a)2)2.
Of course, you have to withdraw the supplement b2/(4a2).
This results in a*((x+b/(2a))2-b2/(4a2)+c/a).
After extension, the a*((x+b/(2a))2+(4ac-b2)/(4a2) results.
Take everything except the binomist out of the clamp:
a*(x+b/(2a))2+(4ac-b2)/(4a).
Compare this with the coordinates of the vertex:
(4ac-b2)/(4a) is the y coordinate of the vertex. You can replace this term with the letter e by defining e=(4ac-b2)/(4a).
If you set the x coordinate of the apex -b/(2a) equal to d, then b/(2a) is the same as -d.
So you come to the vertex form f(x)=a*(x-d)2+e with vertex S (d|e).
Best regards,
Willy
I have not understood any step here and how to apply this to my general parabola or where the derivation comes from or what you expect in the text.I give you examples of what I don’t understand
Then you’re missing a lot of basics. My answer relates to the conversion of the general form of a parabola into the vertex form.
What coordinate point does it look like? “For x=-b/(2a), the vertex of a parabola” and “f(-b/(2a))=a*(-b/(2a))2+b*(-b/(2a))+c” is thus the vertex of a parabola.”
Set the specific values of the given function for a, b and c.
If the equation sounds for example f(x)=3×2-x+4, then a=3, b=-1 and c=4.
Put in, ready.
If the x coordinate of the apex is -b/(2a) and a is equal to 3 and b=-1, then the apex is at x=1/(2*3)=1/6.
Now you can either use and calculate the values of a, b and c in the term for the y coordinate or simply insert x=-1/6 into the function equation, which then comes out to the same.
With Sy (y coordinate of the vertex) equal to (4ac-b2)/(4a) and a=3, b=-1, c=4, you come to coordinate 47/12, namely (4*3*4-1)/12. If you insert x=1/6 in f(x)=3×2-x+4 directly, you will get as a y coordinate of the vertex
3*(1/6)2-1/6+4=47/12, that is exactly the previous result.
In my answer, it was above all about showing how to get to the vertex form of a parabola f(x)=a*(x-d)2+e. Where do the d and the e come from and why is it the same as f(x)=ax2+bx+c?
but I don’t understand your steps
You make a square addition (Note: – x = – 1·x) as usual
Sketch:
simple : There is a not written +1 or -1 ahead
thus halved +0.5 or -0.5
so the x is 1
no: you write 2x for two_x or only x for one_x (1x)
also with x2 = 1×2 or -x2 = -1×2 the 1 is not pushed
.
In the pq formula, at x2 +x+7 the p=+1 and the q=+7
The parabel has a horizontal tangent at the vertex:
p(x) = x^2 – x + 2
p'(x) = 2x – 1 = 0
2x = 1
x = 1/2
p(0.5) = 0.25 – 0.5 + 2 = 1.75
Solution: S(0.5/1,75)
you can enter this, you will see the individual steps:
https://www.mathepower.com/scheitelpunktform.php
if you ask yourself how to come to a half you have to share the prefactor of x by 2 (if I have that right in memory)
I have no prefactor but only x
there is actually 1*x
Okay, thanks
there is 1*x^2-1*x+2, if there is no number in front of x this is one 1 you have a x if you had two there would be 2*x
Where? p(x) = x^2 − x + 2 or if the number is not given before x.