\(y\ =\ \frac{x^3-x^2+x}{6x^2-9x}\)
To find the number of holes, let's factor the numerator and denominator like this: |
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\(y\ =\ \frac{x(x^2-x+1)}{x(6x-9)}\) | |
There is only 1 common factor between the numerator and denominator so there is only 1 hole. |
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a = 1 | |
The vertical asymptotes occur when the denominator after common factors are removed equals zero. |
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6x - 9 = 0 | |
6x = 9 |
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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. |
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c = 0 | |
The degree of the numerator is exactly one greater than the degree of the denominator, so there is an oblique asymptote. |
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d = 1 |
Check: https://www.desmos.com/calculator/vrya3tghg2
a + 2b + 3c + 4d = 1 + 2(1) + 3(0) + 4(1) = 1 + 2 + 4 = 7
x2 + 5x < 6
Subtract 6 from both sides of the inequality.
x2 + 5x - 6 < 0
Let's find what values of x make x2 + 5x - 6 equal 0
x2 + 5x - 6 = 0
Factor the left side. What two numbers add to 5 and multiply to -6 ? -1 and +6
(x - 1)(x + 6) = 0
Set each factor equal to zero and solve for x
x - 1 = 0 or x + 6 = 0 |
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x = 1 x = -6 |
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Since a graph of y = x2 + 5x - 6 is a parabola, we can be sure that |
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the values of x that would make y < 0 fall in one of these two intervals: |
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either the interval (-6, 1) or the interval (-∞, -6) U (1, ∞) |
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To determine which interval is the solution set, let's test a number in both of them. |
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0 is a number in the interval (-6, 1) |
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If x = 0 then y = (0)2 + 5(0) - 6 = -6 |
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And -6 < 0 so we know 0 should be included. |
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2 is a number in the interval (-∞, -6) U (1, ∞) |
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If x = 2 then y = (2)2 + 5(2) - 6 = 8 |
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And 2 > 0 so we know 2 should not be included. |
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So we can be sure that x2 + 5x - 6 < 0 if and only if x is in the interval (-6, 1) |