Consider the surface given by the function f ( x , y ) = 4 x 2 + y 2 − 9 . What is the range of f? Determine any y-intercept(s) of f.For range, I realize that x 2 + y 2 − 9 can be any number, so long as x 2 + y 2 − 9 is positive or equal to zero. However, when looking at f, we know that the denominator cannot equal zero, which means that x 2 + y 2 − 9 can be any number greater than 0. Would this mean that the range of f is ( 0 , ∞ )?For the y-intercept, I know that both x and z need to equal zero. Would this mean that there are no y-intercepts because 0 is not in the range of f?

Question

Consider the surface given by the function   f  (  x  ,  y  )  =      4                  x        2            +              y        2            −      9      . What is the range of   f? Determine any   y-intercept(s) of   f.For range, I realize that             x      2        +          y      2        −    9   can be any number, so long as       x    2    +      y    2    −  9 is positive or equal to zero. However, when looking at   f, we know that the denominator cannot equal zero, which means that             x      2        +          y      2        −    9   can be any number greater than   0. Would this mean that the range of   f is   (  0  ,  ∞  )?For the   y-intercept, I know that both   x and   z need to equal zero. Would this mean that there are no   y-intercepts because   0 is not in the range of   f?

sleuteleni7 · Accepted Answer

There&#039;s one step missing in your computation of the range. Yes, numbers of the form             x      2        +          y      2        −    9   can take any value in   (  0  ,  +  ∞  ). You should have added that it follows from this that then the same thing occurs with       4                  x        2            +              y        2            −      9      .

Consider the surface given by the function f ( x , y ) = 4 <msqr

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