Answer:
Let's complete the square first.
y = x² + 6x + 3
= (x² + 6x + 9) - 6
= (x + 3)² - 6
Therefore, the vertex is (-3, -6) and since the coefficient of (x + 3)² is positive, the vertex is a minimum.
Answer: option d. x = 3π/2Solution: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).
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The answer should be 3.48 X 10^18
6.9
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