1. The functions \(f(x) = x^2\) and \(g(x) = 2^x\) intersect at three points: \((-0.77,0.59)\), \((2,4)\), and \((4,16)\). After the point \((4,16)\) the exponential function \(g(x) = 2^x\) grows more rapidly.

2. The functions \(f(x) = x^{10}\) and \(g(x) = 2^x\) intersect at three points: \((-0.94, 0.52)\), \((1.08, 2.11)\), and another point near \(x \approx 60\). The \(y\) values on both functions at the third point are very very large, but at this point the function \(g(x) = 2^x\) dominates the power function \(f(x) = x^{10}\).

3. For \(a>1\) the exponential function will always dominate the power function.

If the polynomial function has zeros at \(x=2, -3,\) and \(7\) then it can be factored as \(p(x) = C (x-2)(x+3)(x-7)\). To find the leading coefficient \(C\) we can substitute the point \((1,3)\) into the function to get \begin{equation*} 3 = C(1-2)(1+3)(1-7) \quad \implies \quad 3 = C(-1)(4)(-6) \end{equation*} and therefore \( C = \frac{3}{24} = \frac{1}{8}\) and the polynomial function is \begin{equation*} p(x) = \frac{1}{8}(x-2)(x+3)(x-7) \end{equation*}

A degree three (cubic) polynomial can be written in the form \(p(x) = Ax^3 + Bx^2 + Cx + D\). Substituting each point into the polynomial gives four equations in the four unknowns \(A, B, C,\) and \(D\).

Using \((0,5)\) we get \begin{equation*} 5 = A(0)^3 + B(0)^2 + C(0) + D \end{equation*} and hence \(D = 5.\) For the point \((1,-2)\) we get \begin{equation*} -2 = A(1)^3 + B(1)^2 + C(1) + D \end{equation*} and hence \(A+B+C+5 = -2\) so we now know that \(A+B+C = -7\). Next with the point \((3,0)\) we get \begin{equation*} 0 = A(3)^3 + B(3)^2 + C(3) + D \end{equation*} and hence \(27A + 9B + 3C + 5 = 0\) so \(27A + 9B + 3C = -5\). Lastly with the point \((-1,4)\) we get \begin{equation*} 4 = A(-1)^3 + B(-1)^2 + C(-1) + D \end{equation*} so \(-A + B - C + 5 = 4\) which gives us \(-A + B - C = -1\).

There are now three equations in the three remaining unknowns \(A, B\), and \(C\). From here we will use elimination to solve the system of equations. Adding the equation \(A+B+C = -7\) to the equation \(-A+B-C = -1\) gives \(2B = -8\) which tells us that \(B = -4\). Now we can consider just the equations \(A+C = -3\) and \(27A + 3C = 31\). After a bit more algebra we find that \(A = \frac{5}{3}\) and \(C = -\frac{14}{3}\). Therefore the polynomial is \begin{equation*} p(x) = \frac{5}{3} x^3 - 4 x^2 - \frac{14}{3} x + 5. \end{equation*}

1. This function clearly has two roots and the ends show the same behavior. This indicates that this is likely a degree 2 polynomial (a quadratic). Based on the roots \(x=-2\) and \(x=1\) the factors are \((x+2)\) and \((x-1)\) so the polynomial is of the form \(a(x) = C(x+2)(x-1)\). Finally, it appears that the point \((-1,-4)\) is on the curve so \(C=2\) and the polynomial is \begin{equation*} a(x) = 2(x+2)(x-1). \end{equation*}

2. This function clearly has three roots and the ends show the opposite behavior. Since the right-hand side of the function tends down we expect a negative leading coefficient. Based on the roots \(x=-2\), \(x=1\), and \(x=3\) the factors are \((x+2)\), \((x-1)\), and \((x-3)\) and the polynomial takes the form \(b(x) = C(x+2)(x-1)(x-3)\). The point \((-1,-8)\) appears to be on the graph so \(C = -1\) and the polynomial is \begin{equation*} b(x) = -(x+2)(x-1)(x-3). \end{equation*}

3. This function clearly has three roots but the end behavior indicates an even degree polynomial. The root at \(x=1\) is special since we do not cross the \(x\) axis. Hence, the factors are \((x+1)\), \((x-1)\), \((x-1)\), and \((x-2)\). Notice that the \((x-1)\) root is listed twice, and hence the polynomial takes the form \(c(x) = C(x+1)(x-1)^2(x-2)\). The point \((0,-2)\) appears to be on the graph so \(C = 1\) and the polynomial is \begin{equation*} c(x) = (x+1)(x-1)^2(x-2). \end{equation*}

1. For \(f(x) = x^2+3x+2\) and \(g(x) = x-3\) when we form the function \(h(x) = f(x)/g(x)\) we get \(h(x) = \frac{x^2+3x+2}{x-3}\). Notice that \(h(-1) = 0\) and \(h(-2) = 0\) since \(f(-1) = f(-2) = 0\) and \(g(-2) \neq 0\). At \(x=3\), on the other hand, \(g(x) = 0\) and there is a vertical asymptote on the function \(h(x)\). For \(k(x) = g(x) / f(x)\). The vertical asymptotes occur at \(x=-2\) and \(x=-1\) and a zero occurs at \(x=3\).

2. For \(f(x) = x^2-9\) and \(g(x) = x-3\) we see immediately that \(f(x)\) can be factored to \(f(x) = (x-3)(x+3)\) and hence the function \(h(x) = f(x)/g(x)\) simplifies to \(h(x) = x+3\) when \(x\) is not \(3\). When \(x=3\) there is a discontinuity in the graph but this is not a vertical asymptote since the discontinuity can be removed algebraically. For \(k(x) = g(x)/f(x)\) we simplify to \(k(x) = 1/(x+3)\) when \(x\) is not \(3\). Therefore there is a vertical asymptote at \(x=-3\) and a removable discontinuity at \(x=3\).

1. When considering the functions \(f(x) = x^3 + 2x^2 - x + 7\) and \(g(x) = x^2+4x+2\), the cubic function, \(f(x)\), will dominate as \(x \to \infty\). Therefore, as \(x \to \infty\), \(h(x) = f(x)/g(x)\) will go to infinity and \(k(x) = g(x)/f(x)\) will go to zero.

2. When considering the functions \(f(x) = 2x^{4} - 5x^{3} + 8x^{2} - 3x - 1\) and \(g(x) = 3x^{4} - 2x^{2} + 1\) we see that the polynomials are the same degree hence one does not dominate over the other. As \(x\to\infty\) we see that \(h(x) \to \frac{2}{3}\) and \(k(x) \to \frac{3}{2}\).

3. When considering the functions \(f(x) = e^x\) and \(g(x) = x^{10}\) we know that the exponential function will dominate over the power function based on our work in prior activities. Therefore, \(h(x) = f(x)/g(x) \to \infty\) as \(x\to\infty\) and \(k(x) = g(x)/f(x) \to 0\) as \(x \to \infty\).

1. For \(f(x) = \frac{x+3}{5x-2}\) the degrees of the two polynomials are both \(1\) so the numerator and denominator grow at the same rate. The horizontal asymptote will be the ratio of the coefficients: \(f(x) \to \frac{1}{5}\) as \(x \to \infty\). The graph does not cross the horizontal asymptote.

2. In this case where \(g(x) = \frac{x^2+2x-1}{x-1}\) the degree of the numerator is larger than the degree of the denominator so \(g(x) \to \infty\) as \(x\to\infty\) and there is no horizontal asymptote.

3. In this case where \(h(x) = \frac{x+1}{x^2+2x-1}\) the degree of the denominator is larger than the degree of the numerator so \(h(x) \to 0\) as \(x \to \infty\). Therefore, there is a horizontal asymptote at \(y=0\). Observe that \(h(-1) = 0\) so the function's graph does indeed cross the horizontal asymptote.