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

Question

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

istim · Answer

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 · Answer

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

istim · Answer

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

mimi_x3 · Answer

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

mimi_x3 · Answer

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

istim · Answer

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

mimi_x3 · Answer

The minus is inside the bracket.

mimi_x3 · Answer

$$y=((-3)^3+2)(4(3)^2-3)$$

mimi_x3 · Answer

then the y-valule is $-825$

istim · Answer

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 · Answer

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

istim · Answer

Ignore what I said. Thanks. I got it.

mimi_x3 · Answer

(: