Period:
Asymptotes:
Range:
a) y= 2 sec (x) + 1
b) y= -2 csc (x - pi/2)
c) y= 1 + tan (x + pi/4)
d) y= 1/2 + csc (x - pi) + 1
e) y= cot (3x + pi) + 2
Any tips on drawing the graph?
WELL FOR SIN AND AND CSC YOU KNOW THAT THE INTERVAL HAS TO BE BETWEEN 0<X<2PI
SO YOU CAN FOR (B) 0<X-PI/2< 2PI AND THEN SOLVE FOR X. TO SEE WHERE THE GRAPH WILL LIE BETWEEN YOU GET PI/2<X<5PI/2
FOR TAN AND COT YOU KNOW INTERVAL IS PI/2<X<PI/2. FOR COS AND SEC 0<X<PI.
YOU ADDED NUMBER AT THE END MOVES THE WHOLE GRAPH UP OR DOWN WHATEVER NUMBER. AND THE NUMBER IN FRONT OF THE COS,SEC,TAN,COT,CSC,SIN MAKE THE HUMP OR THE HILL STRETCH UP 2 FOR EXAMPLE FOR (A).
HOPE THAT HELPS!
YOU COULD SEE THE GRAPH BY GOING TO GOOGLE AND ENTERING IN WOLFRAM ALPHA A WEBSITE AND ENTER IN THE EQUATION. THEY USUALLY WILL GRAPH IT FOR YOU BUT YOU NEED TO DETERMINE THE PERIOD, AND ETC. STUFF FOR YOUR GRAPH.
To find the period, asymptotes, and range of each of these functions, we can use some basic principles of trigonometric functions. Let's analyze each equation step by step:
a) y = 2 sec(x) + 1:
- Period: The period of the secant function is 2π. Therefore, the period of this function is also 2π.
- Asymptotes: The secant function has vertical asymptotes at intervals of π apart. So this function has vertical asymptotes at x = π/2, x = 3π/2, x = 5π/2, and so on.
- Range: Since the secant function ranges from -∞ to -1 and from 1 to ∞, when we multiply it by 2 and add 1, the range of this function becomes (-∞, -1] U [1, ∞).
b) y = -2 csc(x - π/2):
- Period: The period of the cosecant function is also 2π. So the period of this function remains 2π.
- Asymptotes: The cosecant function has vertical asymptotes at x = kπ, where k is an integer. In this case, we have x - π/2 = kπ, so the asymptotes are at x = kπ + π/2.
- Range: Since the cosecant function ranges from -∞ to -1 and from 1 to ∞, when we multiply it by -2, the range becomes (-∞, -2] U [2, ∞).
c) y = 1 + tan(x + π/4):
- Period: The period of the tangent function is π. So the period of this function remains π.
- Asymptotes: The tangent function has vertical asymptotes at x = (k + 1/2)π, where k is an integer. In this case, we have x + π/4 = (k + 1/2)π, so the asymptotes are at x = (k + 1/2)π - π/4.
- Range: The range of the tangent function is from -∞ to ∞, so when we add 1 to it, the range becomes (-∞, ∞).
d) y = 1/2 + csc(x - π) + 1:
- Period: The cosecant function has a period of 2π. So the period of this function remains 2π.
- Asymptotes: The cosecant function has vertical asymptotes at x = kπ, where k is an integer. In this case, we have x - π = kπ, so the asymptotes are at x = kπ + π.
- Range: Since the cosecant function ranges from -∞ to -1 and from 1 to ∞, when we add 1/2 and 1 to it, the range becomes (-∞, 3/2] U [3/2, ∞).
e) y = cot(3x + π) + 2:
- Period: The cotangent function has a period of π. So when we replace x with (3x + π), the new period becomes π/3.
- Asymptotes: The cotangent function has vertical asymptotes at x = kπ, where k is an integer. In this case, we have 3x + π = kπ, so the asymptotes are at x = (kπ - π)/3.
- Range: The range of the cotangent function is from -∞ to ∞, so when we add 2 to it, the range becomes (-∞, ∞).
Tips for drawing the graphs:
- Start by plotting the asymptotes on the coordinate plane.
- Determine some key points by substituting specific x-values into the function equations.
- Sketch the shape of the curve between asymptotes and pass through the key points.
- Keep in mind the periodicity and symmetry of the functions when extending the graph beyond the given domain.
These tips should help you sketch the graphs of the given trigonometric functions accurately.