1.
Item 1. Which of the following expressions is NOT a rational number?
Solution: A rational number can be written as \(\frac{m}{n}\), where \(m,n\in\mathbb{Z}\) and \(n\ne0\). Option A has denominator \(0\). Option B involves an irrational square root. Option C has irrational \(\sqrt2\) in the denominator. Option D involves \(\sqrt[3]{-16}\), which is not an integer rational value. Therefore all listed expressions are not rational numbers. Answer: E
2.
Item 2. Is \(\frac{16x^{\frac12}-4}{x}\) a rational expression? If not, why not?
Solution: A rational expression must be a ratio of polynomials. Polynomial expressions only allow variables raised to non-negative integer powers. Since \(x^{\frac12}\) is a fractional exponent, the numerator is not a polynomial. Answer: E
3.
Item 3. Determine the non-permissible values of \(\frac{1}{x^2+4}\).
Solution: Non-permissible values occur when the denominator is \(0\). For real \(x\), \(x^2\ge0\), so \(x^2+4>0\). Therefore the denominator never equals \(0\), and there are no real non-permissible values. Answer: A
4.
Item 5. Which function has a vertical asymptote at \(x=-7\)?
Solution: A vertical asymptote at \(x=-7\) comes from the denominator factor \(x+7=0\). The factor must be in the denominator and not cancel with the numerator. Therefore \(y=\frac{x}{x+7}\). Answer: D
5.
Item 7. Which graph represents \(y=\frac{x^2+3x+2}{x^2+6x+8}\)?
Solution: Factor the function: \(y=\frac{x^2+3x+2}{x^2+6x+8}=\frac{(x+2)(x+1)}{(x+2)(x+4)}\). The common factor \(x+2\) creates a point of discontinuity at \(x=-2\). The reduced function is \(y=\frac{x+1}{x+4}\). It has a vertical asymptote at \(x=-4\) and horizontal asymptote \(y=1\). The graph matching these features is Graph C. Answer: C
6.
Item 4. Determine the non-permissible values of \(\frac{2x}{2x^2-7x-4}\).
Solution: Factor the denominator: \(2x^2-7x-4=2x^2-8x+x-4=2x(x-4)+1(x-4)=(2x+1)(x-4)\). Set the denominator not equal to zero: \(2x+1\ne0\Rightarrow x\ne-\frac12\), and \(x-4\ne0\Rightarrow x\ne4\). Answer: E
7.
Item 14. A plane intersects a double-napped cone to form a circle. Assume the plane moves parallel to its original position. Describe what happens to the circle that is formed when the plane moves further away from the vertex.
Solution: When the cutting plane stays parallel and moves further from the vertex, the circular cross section becomes larger. Therefore the radius increases and the circle becomes larger. Answer: C
8.
Item 10. Identify the vertical asymptote(s), if any: \(y=\frac{x^2+5x+6}{2x^2+10x+8}\).
Solution: Factor the denominator: \(2x^2+10x+8=2(x^2+5x+4)=2(x+1)(x+4)\). Vertical asymptotes occur where denominator factors equal zero, so \(x=-1\) and \(x=-4\). Answer: D
9.
Item 6. Identify the intercept(s), if any, in the following graph.
Solution: From the graph, the curves to the left and right do not cross either axis. The central curve crosses the origin \((0,0)\). Therefore the graph has both \(x=0\) and \(y=0\) as intercepts. Answer: B
10.
Item 11. Identify the horizontal asymptote(s), if any: \(y=\frac{24(x+2)(x+3)(x-5)}{-4(x-2)(x-3)}\).
Solution: The numerator has degree \(3\) and the denominator has degree \(2\). When the numerator degree is greater than the denominator degree, there is no horizontal asymptote. Answer: E
11.
Item 17. Which equation represents the following graph?
Solution: From the graph, the center of the ellipse is \((4,-5)\). Therefore the equation must contain \((x-4)^2\) and \((y+5)^2\). The correct equation is \(\frac{(x-4)^2}{4}+\frac{(y+5)^2}{9}=1\). Answer: D
12.
Item 16. What is the equation of a circle with a center at \((0,0)\) and a radius of \(3\)?
Solution: The standard form of a circle centered at \((0,0)\) is \(x^2+y^2=r^2\). Since \(r=3\), \(r^2=9\). Therefore \(x^2+y^2=9\). Answer: B
13.
Item 15. From the following diagram identify the conic formed:
Solution: The plane cuts the cone in a way that forms a parabola. Therefore the conic formed is a parabola. Answer: B
14.
Item 8. Which of the following represents the Domain \(D\) and Range \(R\) for \(y=\frac{x^2-5x+4}{x-4}\)?
Solution: Factor \(x^2-5x+4=(x-1)(x-4)\). The denominator \(x-4\) cancels, so the reduced function is \(y=x-1\), but \(x=4\) is not allowed. At \(x=4\), the missing \(y\)-value is \(y=4-1=3\). Therefore \(D:x\ne4\) and \(R:y\ne3\). Answer: E
15.
Item 9. Describe the end behavior for \(y=-x(x+4)(x-3)(x+2)\).
Solution: The function is degree \(4\) with a negative leading coefficient. A negative quartic has both ends going downward. From left to right, it comes up from \(-\infty\) in Quadrant III and continues down to \(-\infty\) in Quadrant IV. Answer: C
16.
Item 13. Create a rational equation for a function that has vertical asymptotes at \(x=\frac23\) and \(x=-\frac34\).
Solution: For \(x=\frac23\), multiply by \(3\): \(3x=2\), so \(3x-2=0\). For \(x=-\frac34\), multiply by \(4\): \(4x=-3\), so \(4x+3=0\). These factors must be in the denominator and must not cancel. Therefore \(y=\frac{1}{(3x-2)(4x+3)}\). Answer: E
17.
Item 12. Identify any points of discontinuity in \(y=\frac{x^2-5x-6}{3x^2+6x+3}\), if any.
Solution: Factor the numerator and denominator: \(x^2-5x-6=(x-6)(x+1)\) and \(3x^2+6x+3=3(x^2+2x+1)=3(x+1)^2\). The common factor \(x+1\) creates a removable discontinuity at \(x=-1\). Answer: B
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