Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

Given the equations:
[tex]\[ x = t + 4 \][/tex]
[tex]\[ y = t^3 - 12t \][/tex]

A. Horizontal tangents: [tex]\((6, -16), (2, 16)\)[/tex]; Vertical: none

B. Horizontal tangents: none; Vertical: [tex]\((6, -16), (2, 16)\)[/tex]

C. Horizontal tangents: [tex]\((6, 16), (2, 6)\)[/tex]; Vertical: none

D. Horizontal tangents: none; Vertical: [tex]\((6, 16), (2, 16)\)[/tex]

Choose the correct option.


Sagot :

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].