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IsTim
Group Title
Determine an equation for the tangent to each curve at the indicated value.
P(x)=(x^3+2)(4x^23),x=3
 2 years ago
 2 years ago
IsTim Group Title
Determine an equation for the tangent to each curve at the indicated value. P(x)=(x^3+2)(4x^23),x=3
 2 years ago
 2 years ago

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IsTim Group TitleBest ResponseYou've already chosen the best response.1
I've already completed 96% of the question. My final answer is y=149x+4481, but the textbook says it is y=1491x+3648
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
I got \(y=1491x+3648\) as well..probably you did something wrong
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.1
Er...what did you get as your yvalue?
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
\[yy_1 =m(xx_1)\] \[y+825=1491(x3) => y+825=1491x+4473 => y3648=1491x\] \[=> y =1491x+3648\]
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
hm, want me to explain? or you understand it?
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.1
How did you get your yvalue? I did this:\[y=((3)^{3}+2)(4(3)^{2}3)\]
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
The minus is inside the bracket.
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
\[y=((3)^3+2)(4(3)^23)\]
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
then the yvalule is \(825\)
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.1
No, I don't think the minus sign is inside the bracket... My equation is this actually, sorry. \[y=(−(3)^{3}+2)(4(3)^{2}−3)\] I'll try what you said though.
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
Hmm..is it \[P(x) = (x^3+2)(4x^23)\] ?
 2 years ago

IsTim Group TitleBest ResponseYou've already chosen the best response.1
Ignore what I said. Thanks. I got it.
 2 years ago
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