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# f(x)=−10sin(π2x)+4 f(x)=−4sin(πx)+6 f(x)=−4sin(π2x)+6 f(x)=−10sin(πx)+4

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The graph of a sine function has an amplitude of 10, a midline of y = 4, and a period of 2. There is no phase shift. The graph is reflected over the x-axis. What is the equation of the function? f(x)=−10sin(π/2 x)+4 f(x)=−4sin(πx)+6 f(x)=−4sin(π/2 x)+6 f(x)=−10sin(πx)+4

off-topic
Mar 6, 2020

#1
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REFLECTED  sin fxn=    - sinx

amplitude 10   =  - 10 sin x

midline shifted up 4   =   -10 sin x     +4

period = 2                                                     - 10 sin pi x        +4

Mar 6, 2020
edited by ElectricPavlov  Mar 6, 2020
#2
+2

The original function has  the  form

f(x) = Asin (Bx)  + C

A=  the amplitude  = 10

C = the midline shift  =  4

B =  2pi / period    =  2pi  / 2  =  pi

Original  function  is

f(x)  = 10 sin (pi x)  +  4

If  this is  reflected  over  the x axis.....then  f(x)  becomes   -f(x)  =

-10sin (pi x)  -  4

See the  original grah and the reflected  graph here  :  https://www.desmos.com/calculator/gvwssl3gbf   Mar 7, 2020