Consider the graph of \(\frac{x^3-x^2+x}{6x^2-9x}.\) 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.
+9466 \(y\ =\ \frac{x^3-x^2+x}{6x^2-9x}\)
| To find the number of holes, let's factor the numerator and denominator like this: |
|
| \(y\ =\ \frac{x(x^2-x+1)}{x(6x-9)}\) | |
| There is only 1 common factor between the numerator and denominator so there is only 1 hole. |
|
| 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. |
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| 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
