What transformation can you apply to #y=sqrtx# to obtain the graph #y=-sqrt(1/6(x-11)#?

1 Answer
Paramecium
May 20, 2017

See below.

Explanation:

The second graph is basically just a modification of #y = sqrt(x)#.

First, #sqrt(x-11)# is multiplied by a coefficient of #-sqrt(1/6)#. Essentially, this means that the y-value of the second equation is #-(1/6)# times the y-value of #y=sqrt(x)#.

Finally, because 11 is subtracted from #x# within the square root, the graph shifts 11 units to the right.

I have attached the graphs of both equations for reference.
enter image source here

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