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leitkurve y=1-x^2

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scheitelpunkte 25−x²

Solution

Maximum (0, 25)

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y=25−x²

Parabola equation in polynomial form

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

Rewrite y=25−x² in the form y=ax²+bx+c

y=−x²+25

The parabola parameters are:

a=−1, b=0, c=25

x_v=−b2a

x_v=−02(−1)

Vereinfache

x_v=0

Plug in x_v=0 to find the y_v value

y_v=25

Deshalb ist der Parabel Scheitelpunkt

(0, 25)

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

Maximum (0, 25)

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

leitkurve y=1-x^2

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