For \( \frac{(x-4)^{2}}{9}-\frac{(y-2)^{2}}{16}=1 \), \( \begin{array}{ll}\text { (a) Identify the center. } & \text { (b) Identify the vertices. } \\ \begin{array}{ll}\text { (c) Identify the foci. } & \text { (d) Write equations for the asymptotes. } \\ \text { (e) Graph the hyperbola. }\end{array} \\ \text { Give exact answers. Express numbers in simplest form using integers, fractions, or radicals. }\end{array} \)

**(a) Identify the Center** The given equation is \[ \frac{(x-4)^2}{9} - \frac{(y-2)^2}{16} = 1. \] This is in the standard form of a horizontal hyperbola \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1, \] where the center is at \((h, k)\). Comparing, we have \[ h=4 \quad \text{and} \quad k=2. \] Thus, the center is \(\boxed{(4,2)}\). --- **(b) Identify the Vertices** For a horizontal hyperbola, the vertices are located \(a\) units left and right from the center. Here, \[ a^2=9 \quad \Rightarrow \quad a=3. \] So, the vertices are at \[ (4 \pm 3,\,2), \] which gives \[ (7,2) \quad \text{and} \quad (1,2). \] Thus, the vertices are \(\boxed{(7,2)\text{ and }(1,2)}\). --- **(c) Identify the Foci** For a hyperbola, the foci satisfy \[ c^2 = a^2 + b^2. \] We have \[ a^2=9 \quad \text{and} \quad b^2=16, \] so \[ c^2 = 9+16 = 25 \quad \Rightarrow \quad c=5. \] The foci are located \(c\) units left and right of the center (since the hyperbola is horizontal): \[ (4 \pm 5,\,2), \] which gives \[ (9,2) \quad \text{and} \quad (-1,2). \] Thus, the foci are \(\boxed{(9,2)\text{ and }(-1,2)}\). --- **(d) Write Equations for the Asymptotes** For a hyperbola in the form \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1, \] the equations of the asymptotes are \[ y-k = \pm \frac{b}{a}(x-h). \] Substitute \(h=4\), \(k=2\), \(a=3\), and \(b=4\): \[ y-2 = \pm \frac{4}{3}(x-4). \] Thus, the asymptotes are \(\boxed{y-2 = \frac{4}{3}(x-4)}\) and \(\boxed{y-2 = -\frac{4}{3}(x-4)}\). --- **(e) Graph the Hyperbola** 1. **Center:** Plot the point \((4,2)\). 2. **Vertices:** Mark the two vertices at \((7,2)\) and \((1,2)\). 3. **Foci:** Mark the foci at \((9,2)\) and \((-1,2)\). 4. **Asymptotes:** Draw the lines \[ y-2 = \frac{4}{3}(x-4) \quad \text{and} \quad y-2 = -\frac{4}{3}(x-4). \] These lines cross at the center and provide a guide for the shape of the hyperbola. 5. **Hyperbola Branches:** Since the hyperbola is horizontal, the two branches open to the left and right. They approach the asymptotes as \(x\) goes to \(+\infty\) or \(-\infty\). The graph consists of two branches that start near the vertices \((7,2)\) and \((1,2)\) respectively and open outward, approaching the asymptotes. Thus, all required components for the hyperbola have been identified and described.