Solve the equation- 2k^2-5k-18=0
2 answers:
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Answer:
Here's what I get.
Step-by-step explanation:
Question 4
The general equation for a sine function is
y = a sin[b(x - h)] + k
where a, b, h, and k are the parameters.
Your sine wave is
y = 3sin[4(x + π/4)] - 2
Let's examine each of these parameters.
Case 1. a = 1; b = 1; h = 0; k = 0
y = sin x
This is a normal sine curve (the red line in Fig. 1).
(Sorry. I forgot to label the x-axis, but it's always the horizontal axes)
Case 2. a = 3; b = 1; h = 0; k = 0
y = 3sin x
The amplitude changes from 1 to 3.
The parameter a controls the amplitude of the wave (the blue line in Fig. 1).
Case 3. a = 3; b = 1; h = 0; k = 2
y = 3sin x - 2
The graph shifts down two units.
The parameter k controls the vertical shift of the wave (the green line
in Fig. 1).
Case 4. a = 3; b = 4; h = 0; k = 2
y = 3sin(4x) - 2
The period decreases by a factor of four, from 2π to π/2.
The parameter b controls the period of the wave (the purple line in Fig. 2).
Case 5. a = 3; b = 4; h = -π/4; k = 2
y = 3sin[4(x + π/4)] - 2
The graph shifts π/4 units to the left.
The parameter h controls the horizontal shift of the wave (the black dotted line in Fig. 2).
Question 6
y = -1cos[1(x – π)] + 3
Effect of parameters
Refer to Fig. 3.
Original cosine: Solid red line
m = -1: Dashed blue line (reflected across x-axis)
k = 3: Dashed green line (shifted up three units)
b = 1: No change
h = π: Orange line (shifted right by π units)
