well, um, did you try actually drawing the graphs?
x=4√y is the right half of a parabola that opens upward
y=4√x is the upper half of a parabola that opens to the right.
You can think of it as the area in the 1st quadrant, bounded by
y=x^2/16 and x=y^2/16
You have a small lens-shaped region in QI, with vertices at (0,0) and (16,16)
See
http://www.wolframalpha.com/input/?i=plot+&a=*C.plot-_*Calculator.dflt-&a=FSelect_**Plot-.dflt-&f3=x%5E2%2F16+and+4%E2%88%9Ax&f=Plot.plotfunction_x%5E2%2F16+and+4%E2%88%9Ax&f4=x&f=Plot.plotvariable%5Cu005fx&f5=0&f=Plot.plotlowerrange%5Cu005f0&f6=16&f=Plot.plotupperrange%5Cu005f16&f7=0&f=Plot.ymin%5Cu005f0&f8=16&f=Plot.ymax%5Cu005f16
and review your algebra I!
how can you find the bounderies given by the curve x=4sqrt(y) and y=4sqrt(x)? show your solution please. thanks to anyone who will notice this!
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1 answer