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slope of and intercepts f(x)=ln (x-5)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

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∫

ddx

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

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normale di y=4x²−6x, x=2

Solution

y=−110 x+215

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Solution steps

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Trova il punto di tangenza: (2, 4)

Trova la pendenza della y=4x²−6x: 8x−6

Compute the slope at x=2: m=10

Calcola la pendenza della linea perpendicolare: m_p=−110

Trova la retta con la pendenza m=−110 e che passa per (2, 4): y=−110 x+215

y=−110 x+215

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CIAO, Fai una domanda su normale di y=4x²−6x,x=2

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

slope of and intercepts f(x)=ln (x-5)

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