Question

Determine the equation of the tangent to the graph of each curve at the point that corresponds to each value of x.

a) y = (6x-3)(-x^2+2)
I can't believe Im not getting this right. I don't know what I am doing wrong below:

y = -6x^3 + 12x +3x^2 - 6
y' = -18x^2 + 12 + 6x

f(1) = 3
f'(1) = 0

y - 3 = 0(x - 1)
y = 3.

The answer at the back says:
y = 3x-11

Answer 1

the equation for the tangent is: y = m(x-Px)+Py

Answer 2

tangent = (f'(Px))(x-Px)+Py ; where P(x,y) is the point given

Answer 3

So what did I do wrong there?

Answer 4

f(x) = (6x-3)(-x^2+2)

f'(x) = 6(-x^2+2) + (6x-3)(-2x)
f'(x) = -6x^2 +12 -12x^2 +6x

f'(x) = -18x^2 +6x +12

right?

Answer 5

the point (1,3) is given i think

Answer 6

when subbed in x = 1 for y', you get 0.

Answer 7

right, f'(1)=0; which means that the equaion is just y = 3

but are you sure you are looking at the right problem?

Answer 8

my simple mistakes tend to be becasue I left out a part, forgot a sign, or just wrote down the wrong problemm