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Honors Precalculus Limits problem
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XXXX://m.wolframalpha.XXX/XXXXX/?i=+%XXX%5E2-5x%2B4%5D%2F%5Bx%5E2-1%5D+XXXX+-10>x>XX&x=11&y=11</a><br></p> <p>Note the XXXXXXXX blue line is XXXXXXX XXX asymptote XXX XX XXX XXXX XX the XXXXXXXX.</p> <p><XX></p> <p>Edit X:</p> <p>To XXXXX this XX XXXX, you would need to first XXXX the XXXXX</p> <p>Then you XXXX XXX rough shape of the graph (XXXX is XX XXX XXXX XXXXX XX a vertical XXXXXXXXX XX -X XXX a XXXXXXXXXX asymptot XX 1</p> <p>Then you XXXXX compare it XX XXXXX functions, Here XXXX XXXXX most XXXXXX XXXX XXXXXXX XX something XXXX 1/x</p> <p>Finally, check XXX behavior XX either side of the XXXXXXXXX</p> <p>XX we come from the XXXX (XXXXXX below -1, like -X.XX XX we discussed above) the XXXXXXXX XX going XX XX XXXXXXXX because XXX numerator, x-4 is XXXXX to be negative, and XXX XXXXXXXXXXX, x+X is XXXX going to XX negatuve</p> <p>That XX XX XXX x-X XXXXX x&XX;X XX negative</p> <p>XXX x+1 XXXXX x<-X is XXXXXXXX</p> <p>so x-X / x+1 &XX;-1 is positive</p> <p><br></p> <p>XXX XXXXXXXX XX XXX right:</p> <p>XX know XXXX very close from -1 to 4 the function is XXXXX negative</p> <p>XXX XX x=X, XXX function is 0</p> <p>XXXXXXX x-X XX x=X XX 0</p> <p>XXX x-X XX non-XXXX</p> <p>so the function is defined, XXX XXXX</p> <p>So the XXXXXXXX will XXXXX from XXXXXXXX y XXXXXX to positive y values at x=X</p> <p>and XXX both XXXXXXXX, XXX shape will XX XXXXXXX to 1/x (XXXX XX XX say, a partial XXXXX)</p> <p><br></p><p><br></p><p>XXXX X:</p><p>XX find points you would just XXXX a small table, in XXX XXXXXX XX interest</p><p>(x-4)/(x+1)<XX></p><p>so from -XX XX 10 you could XXXXXXXXX XXXX XXXXX(XXXX is the XXXX detail one)</p><table> <tbody><tr> <td>x</td> <td>y</td> </tr> <tr> <td>-10</td> <td>X.555555556</td> </tr> <tr> <td>-9</td> <td>1.XXX</td> </tr> <tr> <td>-X</td> <td>X.XXXXXXXXX</td> </tr> <tr> <td>-X</td> <td>X.XXXXXXXXX</td> </tr> <tr> <td>-6</td> <td>X</td> </tr> <tr> <td>-X</td> <td>2.25</td> </tr> <tr> <td>-4</td> <td>X.666666667</td> </tr> <tr> <td>-X</td> <td>X.5</td> </tr> <tr> <td>-2</td> <td>6</td> </tr> <tr> <td>-1</td> <td>#XXX/0!</td> </tr> <tr> <td>X</td> <td>-4</td> </tr> <tr> <td>1</td> <td>-1.X</td> </tr> <tr> <td>X</td> <td>-0.666666667</td> </tr> <tr> <td>X</td> <td>-X.XX</td> </tr> <tr> <td>X</td> <td>0</td> </tr> <tr> <td>X</td> <td>X.XXXXXXXXX</td> </tr> <tr> <td>6</td> <td>X.XXXXXXXXX</td> </tr> <tr> <td>X</td> <td>X.XXX</td> </tr> <tr> <td>8</td> <td>X.XXXXXXXXX</td> </tr> <tr> <td>X</td> <td>X.X</td> </tr> <tr> <td>10</td> <td>X.XXXXXXXXX</td> </tr></tbody></table><p><br></p> <p>XXXXXX that at x = -1 XX the XXXXXXXX symptom</p><p><XX></p><p>XX, as XXXXX XXXXXX, you know what it looks like roughly (like 1/x)</p><p>but XXXXXXX XXXX that at x=X there XX XXX y=X</p><p>XXX that the vertical asymptote is XX -1 (because x=-X XXXXXXX in dividing by XXXX, so XX such point exists)</p><p>and XXX XXXXXXXXXX asymptote, because it XX a XXXXX of 2 XXXXX order XXXXXXXXXXX, XX the XXXXX XX their co-XXXXXXXXXX (x-X)/(x+1) XXXXX it is XX / XX in XXXX XXXX, so XXX XXXXXXXXXX XXXXXXXXX XX at y=X</p><p>So you can make a XXXXX using y=1 as XXX horinstal portion, XXX x=-1 XX XXX XXXXXXXX XXXXXXX</p><p>and XXX XXXXXXXX XXXX be to the XXX XXXX, and bottom right quadrants, XXXX a XXXXX like XXXX XX X/x</p>">
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