What is the answer for question 4) ? http://postimg.org/image/grjobmz2f/(Reminder): this is 1 question.
The form of the functions will be `y=af(b(x-h))+k` where:
a: is a vertical stretch/compression factor. For sine and cosine this is the amplitude. If a<0 the graph is reflected across the horizontal axis.
b: yields the period through `b=(2pi)/p` (If the normal period is `2pi` ) . (b is a horizontal compression/stretch.) If b<0the graph is reflected across the vertical axis.
h: is the horizontal translation (phase shift).
k: is the vertical translation. For the sine and cosine this gives the midline.
(1) A sine curve with amplitude 2, period `pi` , and a horizontal translation of `pi/3` units to the right.
a=2, `b=(2pi)/pi=2` , `h=pi/3` , and k=0:
(2) A tangent curve reflected over the y-axis, period `3/4` , and translated up 5 units.
a=1, `b=pi/(3/4)=(4pi)/3` --but b<0 because of the reflection so `b=-(4pi)/3` , h=0, k=5.
(3) A cosine curve with period `270^@` , translated down 50 units and reflected over the x-axis.
a=-1 (the reflection), `b=(2pi)/270=pi/135` , h=0, k=-50