1. Exercise 1: Use the vertical line test to determine if each graph represents a function. 2. Formula and rule: The vertical line test says: if any vertical line intersects the graph in more than one point then the relation is not a function, otherwise it is a function. 3. Graph 1 description: a piecewise line that starts below the x-axis, rises crossing the y-axis, then slants down to the right. 4. Graph 1 analysis: every vertical line meets the graph at exactly one point, so it passes the vertical line test. 5. Graph 1 answer: Function. 6. Graph 2 description: two straight lines crossing at the origin forming an X (lines like $y=x$ and $y=-x$). 7. Graph 2 analysis: for many values of $x$ (for example $x=1$) the vertical line meets the graph at two points ($y=1$ and $y=-1$), so it fails the vertical line test. 8. Graph 2 answer: Not a function. 9. Graph 3 description: a semicircular or semi-elliptical arc located to the right of the y-axis (only one $y$ value for each $x$ in its domain). 10. Graph 3 analysis: each vertical line in the domain meets the graph at at most one point, so it passes the vertical line test. 11. Graph 3 answer: Function. 12. Exercise 2: Determine the domain for each relation or function given. 13. Part 1 (table) data: $x\in\{1,4,5\}$ and corresponding $y$ values $-2,8,4$. 14. Part 1 answer (domain): $\{1,4,5\}$. 15. Part 2 (graph): a curve that starts at the origin and goes upward to the right. 16. Part 2 analysis: the curve begins at $x=0$ and continues for larger $x$, so the domain is all real numbers $x$ with $x\ge 0$. 17. Part 2 answer (domain): $[0,\infty)$. 18. Part 3 function: $g(x)=\displaystyle\frac{4}{2x-8}$. 19. Part 3 analysis: the denominator $2x-8$ must not be zero, so $2x-8\neq 0$ which gives $x\neq 4$. 20. Part 3 answer (domain): $(-\infty,4)\cup(4,\infty)$. 21. Part 4 function: $g(x)=7\sqrt{6x-42}$. 22. Part 4 analysis: the radicand must be nonnegative, so $6x-42\ge 0$ which gives $x\ge 7$. 23. Part 4 answer (domain): $[7,\infty)$. 24. Exercise 3: Evaluate the given function expressions at the specified inputs. 25. Part 1: $\rho(x)=3x^{2}-4x+2$ and compute $\rho(-\lambda)$. 26. Part 1 work: substitute $x=-\lambda$ to get $\rho(-\lambda)=3(-\lambda)^{2}-4(-\lambda)+2$. 27. Part 1 simplification: $\rho(-\lambda)=3\lambda^{2}+4\lambda+2$. 28. Part 1 answer: $\rho(-\lambda)=3\lambda^{2}+4\lambda+2$. 29. Part 2: $g(x)=\dfrac{4x^{3}-2}{2x^{7/4}}$ and compute $g(0)$. 30. Part 2 work: substitute $x=0$ gives denominator $2\cdot 0^{7/4}=0$ and numerator $4\cdot0^{3}-2=-2$. 31. Part 2 conclusion: division by zero occurs, so $g(0)$ is undefined. 32. Part 2 answer: $g(0)$ is undefined. 33. Part 3: $h(x)=4x-8+\dfrac{x^{3}}{8}$ and compute $h(3)$. 34. Part 3 work: $h(3)=4\cdot 3-8+\dfrac{3^{3}}{8}$. 35. Part 3 simplification: $h(3)=12-8+\dfrac{27}{8}=4+\dfrac{27}{8}=\dfrac{59}{8}$. 36. Part 3 answer: $h(3)=\dfrac{59}{8}$. 37. Exercise 4: Reflect each given point across the x-axis, the y-axis, and the origin. 38. Reflection rules: across the x-axis $(x,y)\mapsto(x,-y)$, across the y-axis $(x,y)\mapsto(-x,y)$, across the origin $(x,y)\mapsto(-x,-y)$. 39. Point A: $A(3,6)$. 40. A across x-axis: $(3,-6)$. 41. A across y-axis: $(-3,6)$. 42. A across origin: $(-3,-6)$. 43. Point B: $B(-2,8)$. 44. B across x-axis: $(-2,-8)$. 45. B across y-axis: $(2,8)$. 46. B across origin: $(2,-8)$. 47. Point C: $C(6,-12)$. 48. C across x-axis: $(6,12)$. 49. C across y-axis: $(-6,-12)$. 50. C across origin: $(-6,12)$. 51. Point D: $D(-5,-3)$. 52. D across x-axis: $(-5,3)$. 53. D across y-axis: $(5,-3)$. 54. D across origin: $(5,3)$. 55. Exercise 5: For each function give the graph's domain and range and note key features. 56. Part 1 function: $g(x)=(x+2)^{2}$. 57. Part 1 domain: all real numbers, $(-\infty,\infty)$. 58. Part 1 range: since a square is nonnegative and minimum occurs at $x=-2$, range is $[0,\infty)$. 59. Part 1 key feature: vertex at $(-2,0)$ and it opens upward. 60. Part 2 function: $g(x)=(x-1)^{3}$. 61. Part 2 domain: all real numbers, $(-\infty,\infty)$. 62. Part 2 range: all real numbers, $(-\infty,\infty)$, because an odd-degree polynomial is surjective. 63. Part 2 key feature: inflection point at $(1,0)$. 64. Part 3 function: $h(x)=\sqrt{x-4}$ (interpreting the variable as $x$ for domain discussion). 65. Part 3 domain: radicand nonnegative gives $x-4\ge 0$ so domain $[4,\infty)$. 66. Part 3 range: square root is nonnegative so range $[0,\infty)$. 67. Summary: all subparts solved above with domains, ranges, function/not-function decisions, evaluations, and reflected points.