3. For the function \( y=(x-3)^{2}+1 \) find the following: \( \begin{array}{ll}\text { a. X-intercept(s) } & \text { b. Y-intercept(s) }\end{array} \begin{array}{l}\text { c. Type of symmetry } \\ \text { (x-axis, } y \text {-axis, origin, } \\ \text { none) }\end{array} \)

Below is a step-by-step solution for the problem. --- ### Step 1. Find the x-intercept(s) The x-intercept(s) occur when \( y = 0 \). Given the function \[ y = (x-3)^2 + 1, \] set \( y = 0 \): \[ (x-3)^2 + 1 = 0. \] Subtract \(1\) from both sides: \[ (x-3)^2 = -1. \] Since the square of a real number cannot be negative, there are no real values of \(x\) that satisfy this equation. Therefore, the function has **no real x-intercepts**. --- ### Step 2. Find the y-intercept(s) The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the function: \[ y = (0-3)^2 + 1. \] Calculate the square and add: \[ y = 9 + 1 = 10. \] Thus, the y-intercept is at the point: \[ (0, 10). \] --- ### Step 3. Determine the type of symmetry The given function is \[ y = (x-3)^2 + 1. \] This is an upward opening parabola with its vertex at \((3, 1)\). A parabola is symmetric about the vertical line that passes through its vertex. The axis of symmetry is given by: \[ x = 3. \] Thus, the function has **vertical symmetry about the line \(x = 3\)**. --- ### Final Answers - **a. X-intercept(s):** There are no real x-intercepts. - **b. Y-intercept(s):** \((0, 10)\) - **c. Type of symmetry:** Vertical symmetry about the line \(x = 3\).