1.
Item 4. Identify any \(x\)-intercepts in the following equation, if any: \(y=\frac{(x^2-1)(x+3)}{(x^2-9)(x-1)}\).
Solution: Factor the expression: \(x^2-1=(x-1)(x+1)\) and \(x^2-9=(x-3)(x+3)\). Then \(y=\frac{(x-1)(x+1)(x+3)}{(x-3)(x+3)(x-1)}\). The factors \(x+3\) and \(x-1\) cancel, so \(x=-3\) and \(x=1\) are points of discontinuity, not \(x\)-intercepts. The remaining numerator factor is \(x+1\), so the \(x\)-intercept is \(x=-1\). Answer: D
2.
Item 3. Describe the end behavior for the polynomial equation \(y=x(x+4)\).
Solution: Expand \(y=x(x+4)=x^2+4x\). This is a quadratic with positive leading coefficient. A positive quadratic opens upward. Therefore, from left to right, the graph comes down from \(+\infty\) in Quadrant II and continues up to \(+\infty\) in Quadrant I. Answer: B
3.
Item 1. Simplify the rational expression and state the non-permissible values: \(\frac{-64x^3}{14y^3}\).
Solution: Simplify the numerical coefficient: \(\frac{-64}{14}=\frac{-32}{7}\). Therefore \(\frac{-64x^3}{14y^3}=\frac{-32x^3}{7y^3}\). Non-permissible values come from the denominator. Since the denominator is \(7y^3\), we require \(y\ne0\). There is no restriction on \(x\) because \(x\) is only in the numerator. Answer: B
4.
Item 5. Identify any points of discontinuity in the following equation, if any: \(y=\frac{(x^2-1)(x+3)}{(x^2-9)(x-1)}\).
Solution: Factor the expression: \(x^2-1=(x-1)(x+1)\) and \(x^2-9=(x-3)(x+3)\). Then \(y=\frac{(x-1)(x+1)(x+3)}{(x+3)(x-3)(x-1)}\). The common factors \(x+3\) and \(x-1\) cancel, which creates removable discontinuities at \(x=-3\) and \(x=1\). Answer: B
5.
Item 2. Which function has a vertical asymptote at \(x=-7\)?
Solution: A vertical asymptote occurs when the denominator equals \(0\) and the factor does not cancel with the numerator. If the vertical asymptote is \(x=-7\), then \(x+7=0\). Therefore the denominator must contain \(x+7\), and the factor must not cancel. The function \(y=\frac{x}{x+7}\) has vertical asymptote \(x=-7\). Answer: C
6.
Item 6. Which of the following equations has an \(x\)-intercept at \(x=-2\)?
Solution: An \(x\)-intercept at \(x=-2\) requires the numerator to equal \(0\) at \(x=-2\), while the denominator does not equal \(0\) at \(x=-2\). Since \(x=-2\) gives the factor \(x+2=0\), a correct expression would have \(x+2\) in the numerator but not also in the denominator. For \(x^2-2x-8=(x-4)(x+2)\), the factor \(x+2\) appears in the denominator, so \(x=-2\) is not allowed. In option A, \(x+2\) appears in both numerator and denominator, creating a discontinuity rather than an \(x\)-intercept. The other options do not give an \(x\)-intercept at \(x=-2\). Therefore none of these are correct. Answer: B
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