I'm just going to take a dab at this. I haven't done calculus in years.
So you found the equation of the tangent line for your equation at the point (1,2) which is:
y = 5 - 3x
Which is correct and all, but the question seems to be asking for "another tangent", so we are looking for another tangent line that is parallel to:
y = 5 - 3x
So we know that the equation of this line must also have a slope of "-3".
So if it must have a slope of "-3", then it must follow that:
-3 = -3 + 24x - 24x²
I got the above from setting our derivative equation equal to -3.
So if we solve the above equation we get:
24x - 24x² = 0
24x(1 - x) = 0
So "x" must be equal to 0 or 1. (These are the two points that have a slope of -3)
And since we already know about the point x = 1, we are interested in the point x = 0. So from our original equation the point x = 0 has a y-value of 1, therefore we have the point (0,1). Finally, we just need to find the equation of the line that has a slope of -3 and contains the point (0,1).
I'm sure you can do this part on your own. The equation of that line is:
y = -3x +1
or
y = 1 - 3x (as you pointed out)
Whew, that was a nice refresher. :)
I hope that was easy to follow!
