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Questions and Answers
In the inequality $\frac{40}{t+4} - 1 > 0$, what is the significance of making the denominator common when solving for $t$?
In the inequality $\frac{40}{t+4} - 1 > 0$, what is the significance of making the denominator common when solving for $t$?
Making the denominator common allows us to combine the terms into a single fraction, which is essential for determining the intervals where the inequality holds true.
Given a rational function with a vertical asymptote at $x = -1$ and an x-intercept at $x = 2$, explain why the denominator must contain the factor $(x + 1)$ and the numerator must contain the factor $(x - 2)$.
Given a rational function with a vertical asymptote at $x = -1$ and an x-intercept at $x = 2$, explain why the denominator must contain the factor $(x + 1)$ and the numerator must contain the factor $(x - 2)$.
A vertical asymptote at $x = -1$ implies the function is undefined at that point, making $(x + 1)$ a factor of the denominator. An x-intercept at $x = 2$ means the function equals zero at that point, making $(x - 2)$ a factor of the numerator.
How does the horizontal asymptote at $y = -\frac{1}{2}$ influence the coefficients of the numerator and denominator in the rational function $f(x) = -\frac{1}{2} \cdot \frac{x-2}{x+1}$?
How does the horizontal asymptote at $y = -\frac{1}{2}$ influence the coefficients of the numerator and denominator in the rational function $f(x) = -\frac{1}{2} \cdot \frac{x-2}{x+1}$?
The horizontal asymptote at $y = -\frac{1}{2}$ indicates that the ratio of the leading coefficients of the numerator and denominator is equal to $-\frac{1}{2}$.
When solving the inequality $\frac{t-36}{t+4} < 0$, why is it important to consider both the roots of the numerator and the denominator?
When solving the inequality $\frac{t-36}{t+4} < 0$, why is it important to consider both the roots of the numerator and the denominator?
Is it possible to have another function with the same key features (vertical asymptote, x-intercept, horizontal asymptote) as $f(x) = -\frac{1}{2} \cdot \frac{x-2}{x+1}$? Explain your reasoning.
Is it possible to have another function with the same key features (vertical asymptote, x-intercept, horizontal asymptote) as $f(x) = -\frac{1}{2} \cdot \frac{x-2}{x+1}$? Explain your reasoning.
Flashcards
Vertical Asymptote
Vertical Asymptote
A line x = a where a function approaches infinity or negative infinity.
X-intercept
X-intercept
The x-value where the function crosses the x-axis, f(x) = 0.
Horizontal Asymptote
Horizontal Asymptote
A line y = b that a function approaches as x goes to infinity.
Domain of a function
Domain of a function
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Function Equation Formation
Function Equation Formation
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Study Notes
Rational Functions - Unit 2 Test
- General Marking Notations:
- M: Marks for method
- A: Marks for application
- R: Marks for reasoning
- N: Marks awarded when no working is shown, but the answer is correct (no M, A, or R marks)
Question 1
-
Part a):
- Given inequality: (3 / (x - 4)) < 5
- Making the denominator common: (-5x + 23) / (x - 4) < 0
- Simplifying: (x-23/5) / (x-4) <0
- Critical values: x = 4, x = 23/5.
- Solution: x ∈ (-∞, 4) U (23/5, ∞).
-
Part b):
- Given equation: (x² - 8x + 15) / (x² + 5x + 4) ≥ 0
- Factorising: ((x-3)(x-5)) / ((x + 4)(x + 1)) ≥ 0
- Critical values: x = -4, x = -1, x = 3, x = 5.
- Solution set: x ∈ (-∞, -4) U (-1, 3] U [5, ∞)
Question 2.A
- R(t) = (2t) / (t² + 4t):
- Rate is non-zero for all t
- Limit as t approaches ∞: 0.
- Horizontal Asymptote: y = 0
- Vertical asymptote: t = -4
- Behavior around asymptote: R(-4⁻) →-∞ and R(-4⁺) → ∞.
Question 2.B
- Horizontal Asymptote: R(t) = 0
Question 2.C
- Rate greater than 0.05 g/s: (2t) / (t² + 4t) > 0.05
- Finding the values for t where the inequality holds. A solution for t is given, between 0 to 36
- Domain: t ∈ (0, 36)
Question 3
-
Part a):
- Vertical Asymptote at x = -1, so (x + 1) is in the denominator
- x-intercept at x = 2, so (x - 2) is in the numerator
- Horizontal Asymptote at y = 1/2, Meaning that the numerator coefficient divided by the denominator coefficient is 1/2
- Function: f(x) = (x - 2) / (2(x + 1))
-
Part b):
- Yes, it is possible to have a function with similar characteristics by changing the variable restriction.
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