# Caculus assignment

Anonymous

### Question Description

Question 4, 6,7

Itβs for my assignment and I need the full work.

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Borys_S
School: UIUC

The solutions are ready, please ask if something is unclear.docx and pdf files are identical

2. 5π₯ β π₯ 3 π¦ 2 = 4π¦ + 8, tangent line(s) at π₯ = 1.
For π₯ = 1 we obtain 5 β π¦ 2 = 4π¦ + 8, π¦ 2 + 4π¦ + 3 = 0.
There are two solutions, π¦1 = β1, π¦2 = β3.
Differentiate with respect to π₯: 5 β 3π₯ 2 π¦ 2 β 2π₯ 3 π¦π¦ β² = 4π¦ β² .
For π₯ = 1, π¦ = π¦1 = β1 we obtain 5 β 3 + 2π¦ β² = 4π¦ β² , i.e. π¦ β² = 1.
For π₯ = 1, π¦ = π¦2 = β3 we obtain 5 β 27 + 6π¦β² = 4π¦ β² , i.e. π¦ β² = 11.
This way we have two points with known coordinates and tangent slopes.
The first line is π¦ = 1 β (π₯ β 1) β 1 = π β π, the second is π¦ = 11 β (π₯ β 1) β 3 = πππ β ππ.

4. π¦ = (2 β 5π₯ 2 )4 (3π₯ 2 β 8)5 ...

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Anonymous
Solid work, thanks.

Anonymous
The tutor was great. Iβm satisfied with the service.

Anonymous
Goes above and beyond expectations !

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