Given the parametric equations:
[tex]\[ x = t + 4 \][/tex]
[tex]\[ y = t^3 - 12t \][/tex]
We are asked to determine the presence of horizontal and vertical tangents among the following choices:
A. Horizontal tangents: [tex]\((6, -16)\)[/tex] [tex]\((2, 16)\)[/tex], vertical: none
B. Horizontal tangents: none, vertical: [tex]\((6, -16)\)[/tex] [tex]\((2, 16)\)[/tex]
C. Horizontal tangents: [tex]\((6, 16)\)[/tex] [tex]\((2, 6)\)[/tex], vertical: none
D. Horizontal tangents: none, vertical: [tex]\((6, 16)\)[/tex] [tex]\((d, 16)\)[/tex]
E. Horizontal tangents: none, vertical: none
For horizontal tangents, [tex]\(\frac{dy}{dt} = 0\)[/tex]:
[tex]\[ \frac{dy}{dt} = 3t^2 - 12 \][/tex]
Setting the derivative to zero:
[tex]\[ 3t^2 - 12 = 0 \][/tex]
[tex]\[ t^2 = 4 \][/tex]
[tex]\[ t = \pm 2 \][/tex]
For [tex]\( t = 2 \)[/tex]:
[tex]\[ x = 2 + 4 = 6 \][/tex]
[tex]\[ y = 2^3 - 12(2) = 8 - 24 = -16 \][/tex]
Point: [tex]\((6, -16)\)[/tex]
For [tex]\( t = -2 \)[/tex]:
[tex]\[ x = -2 + 4 = 2 \][/tex]
[tex]\[ y = (-2)^3 - 12(-2) = -8 + 24 = 16 \][/tex]
Point: [tex]\((2, 16)\)[/tex]
Thus, the points of horizontal tangency are [tex]\((6, -16)\)[/tex] and [tex]\((2, 16)\)[/tex].
For vertical tangents, [tex]\(\frac{dx}{dt} = 0\)[/tex]:
[tex]\[ \frac{dx}{dt} = 1 \][/tex]
Since [tex]\(\frac{dx}{dt}\)[/tex] is never zero, there are no vertical tangents.
Thus, we have:
- Horizontal tangents: [tex]\((6, -16)\)[/tex], [tex]\((2, 16)\)[/tex]
- Vertical tangents: none
Given these findings, the correct choice is:
[tex]\[
\text{A. Horizontal tangents: } (6, -16) \text{, } (2, 16), \text{ vertical: none}
\][/tex]
Therefore, the answer is [tex]\(\boxed{1}\)[/tex].
