1. **Problem:** For the quadratic function $y = 3(x - 1)^2 + 4$, determine the number of x-intercepts, the equation of the axis of symmetry, and the domain and range. 2. **Formula and rules:** - The general form of a quadratic function is $y = a(x - h)^2 + k$ where $(h,k)$ is the vertex. - The axis of symmetry is the vertical line $x = h$. - The number of x-intercepts depends on the discriminant or by checking if the parabola crosses the x-axis. - The domain of any quadratic function is all real numbers: $(-\infty, \infty)$. - The range depends on the vertex and whether the parabola opens up ($a > 0$) or down ($a < 0$). 3. **Step 1: Identify vertex and axis of symmetry** - Here, $a = 3$, $h = 1$, $k = 4$. - Axis of symmetry: $x = 1$. 4. **Step 2: Find x-intercepts by setting $y=0$** $$0 = 3(x - 1)^2 + 4$$ $$3(x - 1)^2 = -4$$ $$ (x - 1)^2 = \frac{-4}{3}$$ 5. **Step 3: Analyze the equation** - Since $(x - 1)^2$ equals a negative number, there are no real solutions. - Therefore, the parabola does not cross the x-axis. - Number of x-intercepts = 0. 6. **Step 4: Domain and range** - Domain: all real numbers, $(-\infty, \infty)$. - Since $a=3 > 0$, parabola opens upward. - Vertex is minimum point at $y=4$. - Range: $[4, \infty)$. **Final answers:** - Number of x-intercepts = 0 - Equation of axis of symmetry = $x = 1$ - Domain = $(-\infty, \infty)$ - Range = $[4, \infty)$