1.
Which of the following represents the domain \(D\) and range \(R\) for \(y=\frac{x^2-5x+4}{x-4}\)?
Factor the numerator: \(x^2-5x+4=(x-1)(x-4)\). Then \(y=\frac{(x-1)(x-4)}{x-4}\), so \(x\ne4\). After cancellation, \(y=x-1\). At the removed point \(x=4\), the corresponding \(y\)-value would be \(3\), so \(y\ne3\). Answer: C
2.
Identify any points of discontinuity in \(y=\frac{x^3+6x^2+11x+6}{x^3-6x^2+11x-6}\), if any.
Factor the numerator: \(x^3+6x^2+11x+6=(x+1)(x+2)(x+3)\). Factor the denominator: \(x^3-6x^2+11x-6=(x-1)(x-2)(x-3)\). There are no common factors, so there are no removable points of discontinuity. Answer: E
3.
Determine the non-permissible value(s) of the variable in the expression: \(\frac{2q-7}{q}\).
The denominator cannot equal zero. Since the denominator is \(q\), set \(q\ne0\). Therefore, the non-permissible value is \(q=0\). Answer: C
4.
Identify any \(x\)-intercepts in \(y=\frac{x^2-5x-6}{3x^2+6x+3}\), if any.
The \(x\)-intercepts come from the numerator, as long as the denominator is not zero. Factor the numerator: \(x^2-5x-6=(x-6)(x+1)\). Factor the denominator: \(3x^2+6x+3=3(x+1)^2\). Since \(x=-1\) makes the denominator zero, it is not an \(x\)-intercept. The only \(x\)-intercept is \(x=6\). Answer: A
5.
Which equation best describes the given graph?
The graph is a reflected reciprocal-type graph, so it has a negative sign. It has a vertical asymptote at \(x=5\), which gives a factor \((x-5)\) in the denominator. It also has a removable discontinuity at \(x=2\), which means \((x-2)\) appears in both the numerator and denominator. Therefore, the equation is \(y=-\frac{x-2}{(x-5)(x-2)}\). Answer: A
6.
Which graph represents \(y=\frac{x^2-1}{-x^2+4x-3}\)?
Factor: \(x^2-1=(x-1)(x+1)\), and \(-x^2+4x-3=-(x-1)(x-3)\). The common factor \(x-1\) creates a hole at \(x=1\). After cancellation, \(y=-\frac{x+1}{x-3}\). There is a vertical asymptote at \(x=3\) and a horizontal asymptote at \(y=-1\). This matches Graph A. Answer: A
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