Nice, already in vertex form
y=a(x-h)^2+k
(h,k) is vertex
therfor since (-3,6) is vertex
we are looking for something like
y=a(x-(-3))^2+6 simplified to
y=a(x+3)^2+6
A is ansre
Answer:
y = 4 sin(½ x) − 3
Step-by-step explanation:
The function is either sine or cosine:
y = A sin(2π/T x) + C
y = A cos(2π/T x) + C
where A is the amplitude, T is the period, and C is the midline.
The midline is the average of the min and max:
C = (1 + -7) / 2
C = -3
The amplitude is half the difference between the min and max:
A = (1 − -7) / 2
A = 4
The maximum is at x = π, and the minimum is at x = 3π. The difference, 2π, is half the period. So T = 4π.
Plugging in, the options are:
y = 4 sin(½ x) − 3
y = 4 cos(½ x) − 3
Since the maximum is at x = π, this must be a sine wave.
y = 4 sin(½ x) − 3
(-2+8i)x(3-10i)
-2 x 3 - 2 x (-10i) +8i x 3- 8i x 10i
-6+20i+24i - 80i ^2
-6+20i+24i - 80 x (-1)
-6+20i+24i + 80
Answer: 74+44i
