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points of d'intersection 1/(1+x^2)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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axe y=x²−6x+8

Solution

x=3

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y=x²−6x+8

Equation standard d'une parabole

4p(y−k)=(x−h)² est l'équation standard pour une parabole tournée de haut en bas avec un sommet en (h, k),
et de longueur focale |p|

Récrire y=x²−6x+8 dans la forme standard

4· 14 (y−(−1))=(x−3)²

Par conséquent les propriétés de la parabole sont :

(h, k)=(3, −1), p=14

La parabole est de la forme 4p(y−k)=(x−h)² et elle est symétrique autour de l'axe y
L′axe de symétrie est la droite parallèle à l′axe }y qui coupe le sommet :

x=3

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

points of d'intersection 1/(1+x^2)

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