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

1. 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

2. Mimi_x3

I got \(y=-1491x+3648\) as well..probably you did something wrong

3. IsTim

Er...what did you get as your y-value?

4. Mimi_x3

\[y-y_1 =m(x-x_1)\] \[y+825=-1491(x-3) => y+825=-1491x+4473 => y-3648=-1491x\] \[=> y =-1491x+3648\]

5. Mimi_x3

hm, want me to explain? or you understand it?

6. IsTim

How did you get your y-value? I did this:\[y=(-(3)^{3}+2)(4(3)^{2}-3)\]

7. Mimi_x3

The minus is inside the bracket.

8. Mimi_x3

\[y=((-3)^3+2)(4(3)^2-3)\]

9. Mimi_x3

then the y-valule is \(-825\)

10. 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.

11. Mimi_x3

Hmm..is it \[P(x) = (-x^3+2)(4x^2-3)\] ?

12. IsTim

Ignore what I said. Thanks. I got it.

13. Mimi_x3

(: