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eixos y=3x^2-30x+77

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

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ddx

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

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normal de 4x², at x=1

Solution

y=−18 x+338

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Encontrar o ponto de tangência: (1, 4)

Calcular a inclinação de y=4x²: 8x

Compute the slope at x=1: m=8

Calcular a inclinação de uma reta perpendicular: m_p=−18

Encontrar a reta com inclinação m=−18 que passa por (1, 4): y=−18 x+338

y=−18 x+338

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

eixos y=3x^2-30x+77

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eixos y=3x^2-30x+77

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