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What is the answer for question 4) ? http://postimg.org/image/grjobmz2f/(Reminder):...

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lkballer24 | Student, Grade 11 | (Level 1) Valedictorian

Posted August 2, 2013 at 10:21 AM via web

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What is the answer for question 4) ?

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted August 2, 2013 at 1:11 PM (Answer #1)

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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:

`y=2sin(2(x-pi/3))`

The graph:

(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.

`y=tan((-4pi)/3x)+5`

The graph:

(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

`y=-cos(pi/135x)-50`

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