The region D above lies between the graphs of y = - 5 – (x – 2) and 1 9 + -(x – 0)°. It can be described in two ways. 9 y = 1. If we visualize the region having "top" and "bottom" boundaries, express each as functions of x and provide the interval of x-values that covers the entire region. "top" boundary g2 (x) = | -5 – (x – 2)² of "bottom" boundary g1(x) = | -9+ (x – 0)3 interval of values that covers the region = 2. If we visualize the region having "right" and "left" boundaries, then the "right" boundary must be defined piece-wise. Express each as functions of y for the provided intervals of y-values that covers the entire region. For – 6 < y < - 5 the "right" boundary as a piece-wise function f2(y) = For – 9 < y < – 6 the "right" boundary f2(y) = %3D For – 9 < y < 5 the "left" boundary f1(y) =

Describe how the graph of y = −3f[2(x + 5)] − 4 can be obtained from the graph of f(x) = x3 . Be sure to fully describe the transformations.

How do the graphs of y=1/x and y=5/x+6

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