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Honors Precalculus Limits problem
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XXXX://m.XXXXXXXXXXXX.com/XXXXX/?i=+%5Bx%XXX-5x%2B4%5D%XX%XXX%XXX-1%5D+from+-XX&XX;x&XX;10&x=XX&y=11</a><XX></p> <p>Note XXX vertical XXXX XXXX is marking the asymptote XXX is XXX XXXX of the XXXXXXXX.</p> <p><XX></p> <p>Edit 2:</p> <p>XX graph this by hand, you XXXXX XXXX to XXXXX note XXX above</p> <p>Then you XXXX XXX XXXXX shape XX the graph (that XX to say XXXX there is a XXXXXXXX XXXXXXXXX XX -1 and a horizontal asymptot at 1</p> <p>XXXX you XXXXX compare it to other XXXXXXXXX, Here this would most likely XXXX XXXXXXX to XXXXXXXXX XXXX 1/x</p> <p>XXXXXXX, check the behavior on either XXXX of the XXXXXXXXX</p> <p>XX XX XXXX from XXX XXXX (XXXXXX below -1, like -1.01 XX we XXXXXXXXX XXXXX) XXX XXXXXXXX is going to be XXXXXXXX XXXXXXX the XXXXXXXXX, x-X is XXXXX to XX negative, XXX the XXXXXXXXXXX, x+1 is XXXX going XX be XXXXXXXX</p> <p>XXXX XX to say x-4 where x<X XX negative</p> <p>and x+X XXXXX x&XX;-1 XX negative</p> <p>so x-X / x+X <-1 is XXXXXXXX</p> <p><XX></p> <p>XXX XXXXXXXX XX the right:</p> <p>we XXXX that very close XXXX -1 XX 4 the XXXXXXXX is still XXXXXXXX</p> <p>and XX x=X, XXX function XX X</p> <p>because x-4 XX x=4 is X</p> <p>and x-1 XX non-zero</p> <p>so XXX function is XXXXXXX, and zero</p> <p>So XXX function XXXX cross from XXXXXXXX y values XX XXXXXXXX y XXXXXX XX x=X</p> <p>and for XXXX segments, the XXXXX XXXX be similar to 1/x (that is to XXX, a partial curve)</p> <p><XX></p><p><br></p><p>Edit 3:</p><p>XX find points you XXXXX just make a small table, in the region of interest</p><p>(x-4)/(x+1)<br></p><p>so XXXX -10 to 10 you could XXXXXXXXX XXXX point(this XX XXX XXXX detail XXX)</p><table> <tbody><tr> <td>x</td> <td>y</td> </tr> <tr> <td>-XX</td> <td>X.XXXXXXXXX</td> </tr> <tr> <td>-X</td> <td>1.XXX</td> </tr> <tr> <td>-8</td> <td>1.XXXXXXXXX</td> </tr> <tr> <td>-7</td> <td>X.833333333</td> </tr> <tr> <td>-X</td> <td>X</td> </tr> <tr> <td>-X</td> <td>X.XX</td> </tr> <tr> <td>-X</td> <td>X.666666667</td> </tr> <tr> <td>-3</td> <td>X.5</td> </tr> <tr> <td>-X</td> <td>6</td> </tr> <tr> <td>-1</td> <td>#XXX/X!</td> </tr> <tr> <td>X</td> <td>-4</td> </tr> <tr> <td>1</td> <td>-X.X</td> </tr> <tr> <td>X</td> <td>-X.XXXXXXXXX</td> </tr> <tr> <td>3</td> <td>-X.XX</td> </tr> <tr> <td>4</td> <td>0</td> </tr> <tr> <td>X</td> <td>0.XXXXXXXXX</td> </tr> <tr> <td>X</td> <td>0.285714286</td> </tr> <tr> <td>X</td> <td>X.375</td> </tr> <tr> <td>X</td> <td>0.444444444</td> </tr> <tr> <td>9</td> <td>0.5</td> </tr> <tr> <td>XX</td> <td>0.XXXXXXXXX</td> </tr></tbody></table><p><XX></p> <p>Noting XXXX at x = -X is XXX XXXXXXXX XXXXXXX</p><p><br></p><p>Or, as XXXXX before, you know XXXX it looks XXXX roughly (like X/x)</p><p>but shifted such XXXX XX x=4 there is XXX y=0</p><p>XXX that the XXXXXXXX asymptote is XX -1 (because x=-1 results in dividing XX XXXX, so no such point exists)</p><p>and the XXXXXXXXXX XXXXXXXXX, because it XX a ratio XX X first order polynomials, is XXX ratio of their co-XXXXXXXXXX (x-X)/(x+X) XXXXX it XX XX / 1x in each XXXX, so the XXXXXXXXXX asymptote is at y=1</p><p>XX you can XXXX a cross using y=1 as XXX XXXXXXXXX portion, and x=-1 XX XXX XXXXXXXX XXXXXXX</p><p>XXX the function XXXX be to the top XXXX, and XXXXXX XXXXX quadrants, with a XXXXX like that of 1/x</p>">
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