A vertical line has a(n) ________ slope. a. undefined b. negative c. zero d. positive
1 answer:
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Answer:
Step-by-step explanation:
From the graph attached,
Coordinates of the vertices are,
Q(1, 3), R(3, -3), S(0, -2) and T(-2, 1)
Following the rule of translation by 3 units to the right and 2 units down
(x, y) → (x+3, y-2)
Q(1, 3) → Q''(4, 1)
R(3, -3) → R"(6, -5)
S(0, -2) → S"(3, -4)
T(-2, 1) → T"(1, -1)
Following rule (rotation of a point by 180° about the origin) will give the image points,
(x, y) → (-x, -y)
Q"(4, 1) → Q'(-4, -1)
R"(6, -5) → R'(-6, 5)
S"(3, -4) → S'(-3, 4)
T"(1, -1) → T'(-1, 1)
Answer: option d. x = 3π/2
Solution:
function y = sec(x)
1) y = 1 / cos(x)
2) When cos(x) = 0, 1 / cos(x) is not defined
3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...
4) limit of sec(x) = lim of 1 / cos(x).
When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.
So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).
Answer: 3π/2
The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).
Solution:
function y = sec(x)
1) y = 1 / cos(x)
2) When cos(x) = 0, 1 / cos(x) is not defined
3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...
4) limit of sec(x) = lim of 1 / cos(x).
When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.
So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).
Answer: 3π/2
The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).