Answer:
4. 2 sin(πx/180 - 225π/180) + 1
5. 8 sin(x-30º) + 2
Step-by-step explanation:
For question 4, the range of the function is [-1, 3]. The range of a normal sine function would be [-1, 1]. Since the range of the function is twice as large, the value of ‘a’ would be 2 since the function has been vertically stretched by a factor of 2. The function has also shifted 1 unit up, leaving the value of ‘c’ as 1. The distance between a period shown (the troughs) is 495-135 which is 360. The distance between two periods in a normal sine function would be 2π. This means that the graph was horizontally shrunken by a factor of π, but horizontally stretched by a factor of 180 as well. In a normal sine function, when x=0, the function is in the middle of the function’s possible y-values and increases. This happes at x=225, meaning that the function was shifted by 225 units. Putting all of this together, you get 2 sin(π(x-225)/180) + 1. Rewriting this out, you get
y = 2 sin(πx/180 - 225π/180) + 1
For question 5, the range of the graph varies 16 units at most. The normal sine function’s range varies 2 units at most. This means that this function is a sine functon stretched by a vertical factor of 8, leaving ‘a’ to equal 8. Taking the average of the highest and lowest y-values (-6 and 10) would give the midline, being 2. In a normal sine function, this would be 0, meaning that the function was shifted vertically 2 units up. This results in y = 8 sin(x-d) + 2
If x = 30º when y = 2, this means that 8 sin(x-d) = 0
sin(x-d) = 0
sin(30º-d) = 0
d is therefore 30º because sin (0) = 0
The equation is thus y = 8 sin(x-30º) + 2
There are an infinite amount of answers where ‘a’ could be -8, or where d could vary in increments of +/- 180º