Consider the graph of x3−x2+x6x2−9x. Let a be the number of holes in the graph, b be the number of vertical asympotes, c be the number of horizontal asymptotes, and d be the number of oblique asymptotes. Find a+2b+3c+4d.
+9489 y = x3−x2+x6x2−9x
| To find the number of holes, let's factor the numerator and denominator like this: |
|
| y = x(x2−x+1)x(6x−9) | |
| There is only 1 common factor between the numerator and denominator so there is only 1 hole. |
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| a = 1 | |
| The vertical asymptotes occur when the denominator after common factors are removed equals zero. |
|
| 6x - 9 = 0 | |
| 6x = 9 |
|
| x = 3/2 | |
| There is only 1 vertical asymptote. |
|
| b = 1 | |
| Since the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes. |
|
| c = 0 | |
| The degree of the numerator is exactly one greater than the degree of the denominator, so there is an oblique asymptote. |
|
| d = 1 |
Check: https://www.desmos.com/calculator/vrya3tghg2
a + 2b + 3c + 4d = 1 + 2(1) + 3(0) + 4(1) = 1 + 2 + 4 = 7
