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bereich 3^x

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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scheitelpunkte x²−2x−3

Solution

Minimum (1, −4)

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Find vertex using polynomial form

Find vertex using parabola form

Find vertex using vertex form

Find vertex using averaging the zeros

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y=x²−2x−3

Parabola equation in polynomial form

The vertex of an up-down facing parabola of the form y=ax²+bx+c is x_v=−b2a

The parabola parameters are:

a=1, b=−2, c=−3

x_v=−b2a

x_v=−(−2)2· 1

Vereinfache

x_v=1

Plug in x_v=1 to find the y_v value

y_v=−4

Deshalb ist der Parabel Scheitelpunkt

(1, −4)

Wenn a<0, dann entspricht der Scheitpunkt dem Maximalwert
Wenn a>0, dann entspricht der Scheitelpunkt dem Minimalwert
a=1

Minimum (1, −4)

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Lösen durch: Find vertex using parabola form

Lösen durch: Find vertex using vertex form

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

bereich 3^x

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