1.
Item 2. Which of the following accurately represents the equations of the horizontal and vertical asymptotes of \(y=\frac1{x-2}+3\)?
Solution: In the form \(y=\frac1{x-h}+k\), the vertical asymptote is \(x=h\) and the horizontal asymptote is \(y=k\). Hence \(x=2,\ y=3\). Answer: A
2.
Item 1. Simplify the rational expression and state the non-permissible values: \(\frac{112x^5}{32}\).
Solution: Simplify \(\frac{112}{32}=\frac72\). There is no variable in the denominator, so there are no non-permissible values. Final answer: \(\frac{7x^5}{2}\). Answer: C
3.
Item 6. Which equation best describes the graph?
Solution: The graph is a reflection of \(y=\frac1x\) across the \(x\)-axis, has a vertical asymptote at \(x=5\), and a removable discontinuity at \(x=2\). Therefore the correct equation is the one represented by Graph D. Answer: D
4.
Item 3. Which of the following represents the Domain and Range for \(y=\frac{x^2-5x+4}{x-4}\)?
Solution: Factor the numerator: \(x^2-5x+4=(x-1)(x-4)\). Cancel \(x-4\) to get \(y=x-1\), with \(x\ne4\). The missing point is \((4,3)\), so \(D:x\ne4,\ R:y\ne3\). Answer: C
5.
Item 4. Identify any points of discontinuity in \(y=\frac{x^3+6x^2+11x+6}{x^3-6x^2+11x-6}\).
Solution: Numerator factors to \((x+1)(x+2)(x+3)\). Denominator factors to \((x-1)(x-2)(x-3)\). There are no common factors, so there are no removable discontinuities (holes). Answer: E
6.
Item 5. Identify any \(x\)-intercepts in \(y=\frac{x^2-5x-6}{3x^2+6x+3}\).
Solution: Factor numerator: \(x^2-5x-6=(x-6)(x+1)\). Denominator: \(3(x+1)^2\). Since \(x=-1\) is not in the domain, it is a hole, not an intercept. The remaining intercept is \(x=6\). Answer: C
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