IsTim
  • IsTim
Determine an equation for the tangent to each curve at the indicated value. P(x)=(-x^3+2)(4x^2-3),x=3
Mathematics
jamiebookeater
  • jamiebookeater
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IsTim
  • IsTim
I've already completed 96% of the question. My final answer is y=-149x+4481, but the textbook says it is y=-1491x+3648
Mimi_x3
  • Mimi_x3
I got \(y=-1491x+3648\) as well..probably you did something wrong
IsTim
  • IsTim
Er...what did you get as your y-value?

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Mimi_x3
  • Mimi_x3
\[y-y_1 =m(x-x_1)\] \[y+825=-1491(x-3) => y+825=-1491x+4473 => y-3648=-1491x\] \[=> y =-1491x+3648\]
Mimi_x3
  • Mimi_x3
hm, want me to explain? or you understand it?
IsTim
  • IsTim
How did you get your y-value? I did this:\[y=(-(3)^{3}+2)(4(3)^{2}-3)\]
Mimi_x3
  • Mimi_x3
The minus is inside the bracket.
Mimi_x3
  • Mimi_x3
\[y=((-3)^3+2)(4(3)^2-3)\]
Mimi_x3
  • Mimi_x3
then the y-valule is \(-825\)
IsTim
  • IsTim
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.
Mimi_x3
  • Mimi_x3
Hmm..is it \[P(x) = (-x^3+2)(4x^2-3)\] ?
IsTim
  • IsTim
Ignore what I said. Thanks. I got it.
Mimi_x3
  • Mimi_x3
(:

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