1.
Item 4. Identify the horizontal asymptote(s), if any. \(y=\frac{24(x+2)(x+3)(x-5)}{-4(x-2)(x-3)}\).
Solution: Compare the degree of the numerator and denominator. The numerator \(24(x+2)(x+3)(x-5)\) has degree \(3\), and the denominator \(-4(x-2)(x-3)\) has degree \(2\). For a rational function, if the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. Answer: C
2.
Item 1. Determine the non-permissible values of \(\frac{2x}{2x^2-7x-4}\).
Solution: Non-permissible values occur when the denominator equals \(0\). Factor \(2x^2-7x-4\). Since \(2\cdot(-4)=-8\), use \(-8\) and \(1\): \(2x^2-7x-4=2x^2-8x+x-4=2x(x-4)+1(x-4)=(2x+1)(x-4)\). Set each factor not equal to \(0\): \(2x+1\ne0\Rightarrow x\ne-\frac12\), and \(x-4\ne0\Rightarrow x\ne4\). Answer: E
3.
Item 2. Identify the intercept(s), if any, in the following graph.
Solution: From the graph, the curves on the left and right do not cross either axis. The central curve crosses the origin \((0,0)\). Therefore there are two intercepts: \(x=0\) and \(y=0\). Answer: A
4.
Item 5. Identify any points of discontinuity in the equation \(y=\frac{x^3+6x^2+11x+6}{x^3-6x^2+11x-6}\), if any.
Solution: Factor the denominator \(x^3-6x^2+11x-6=(x-1)(x-2)(x-3)\). The denominator cannot equal \(0\), so \(x\ne1,2,3\). The numerator factors as \(x^3+6x^2+11x+6=(x+1)(x+2)(x+3)\), so there are no common factors with the denominator. Therefore the domain restriction is \(D=\{x\mid x\in\mathbb{R},x\ne1,2,3\}\). Answer: B
5.
Item 3. Which statement is true about the function \(y=-\frac{x}{x^2-x}\)?
Solution: Factor the denominator: \(x^2-x=x(x-1)\). Then \(y=-\frac{x}{x(x-1)}\). The common factor \(x\) cancels, so there is a removable discontinuity at \(x=0\). The reduced function is \(y=-\frac{1}{x-1}\). Evaluate at \(x=0\): \(y=-\frac{1}{0-1}=1\). Therefore the point of discontinuity is \((0,1)\). Answer: C
6.
Item 6. Create a possible rational equation for a function that has vertical asymptotes at \(x=-\frac89\) and \(x=\frac57\).
Solution: Vertical asymptotes occur where denominator factors equal \(0\). For \(x=-\frac89\), multiply by \(9\): \(9x=-8\), so \(9x+8=0\). For \(x=\frac57\), multiply by \(7\): \(7x=5\), so \(7x-5=0\). Therefore the denominator must contain \((9x+8)(7x-5)\). The numerator and vertical shift do not change the vertical asymptotes, so a possible equation is \(y=\frac{-5}{(9x+8)(7x-5)}+3\). Answer: B
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