How do you graph f ( x ) = 3 x 2 ( x + 5 ) using holes, vertical and horizontal asymptotes, x and y intercepts?

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How do you graph             f              (        x        )              =          3                        x          2                          (          x          +          5          )                     using holes, vertical and horizontal asymptotes, x and y intercepts?

Leon Webster · Accepted Answer

For vertical asymptotes, look at the denominator.             x      2              (      x      +      5      )        ≠    0   because the graph will be undefined            x      2              (      x      +      5      )        ≠    0   means that the vertical asymptotes are x=0 and x=−5 when solving for xFor the horizontal asymptote, look at the degree of the numerator and denominatorIf the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is y=0.Or you can think of it as if you put some numbers into your f(x), you will notice that             x      2              (      x      +      5      )       will be a lot bigger than 3. Hence, when you divide a small number by a big number,             3                        x          2                          (          x          +          5          )                      →    0  For your x and y intercepts,sub y=0 for x intercepts,      0    =          3                        x          2                          (          x          +          5          )                    0=3 which isn&#039;t true so no x-interceptssub x=0 for y intercepts which cannot occur since x=0 is an asymptoteTherefore, there are no interceptsBelow is the graph. You can see that the endpoints of the graphs approaches the asymptotes y=0, x=0 and x=−5graph{3/(x^2(x+5) [-10, 10, -5, 5]}

How do you graph f(x)=3/x^2(x+5) using holes, vertical and horizontal asymptotes, x and y intercepts?

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