vertices 3a^2+1-(4+2a^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

simplify solve for expand factor rationalize

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focos y=2x²−5

Solution

(0, −398 )

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Foco de uma parábola

Uma parábola é o espaço de pontos tal que a distância a um ponto (o foco) equivale à distância a uma reta (a diretriz)

Equação geral da parábola

4p(y−k)=(x−h)² é a equação geral da parábola de eixo vertical, com vértice em (h, k),
e distância focal |p|

Reescrever y=2x²−5 com a forma da equação geral da parábola: 4· 18 (y−(−5))=(x−0)²

(h, k)=(0, −5), p=18

A parábola é simétrica ao redor do eixo y (ordenadas) e, portanto, o foco estabelece uma distância p do centro (0, −5) ao longo do eixo y (ordenadas)

(0, −5+p)

=(0, −5+18 )

Simplificar

(0, −398 )

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

vertices 3a^2+1-(4+2a^2)

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