The given problem involves a quadratic function $f(x) = 2(x - 1)^2 + 5$. It asks for the following:

Part (a)

Sketch the graph of $f(x)$ with all steps:

To sketch the graph, let's break it down:

1. Standard form of the function:
   The given function $f(x) = 2(x - 1)^2 + 5$ is in vertex form: $f(x) = a(x - h)^2 + k,$ where $(h, k)$ is the vertex, and $a = 2$ determines the direction and steepness of the parabola. Here:

   • Vertex: $(h, k) = (1, 5)$,
   • $a = 2$: The parabola opens upward and is vertically stretched.

2. Axis of symmetry: The axis of symmetry is the vertical line passing through the vertex, i.e., $x = 1$.

3. Key points: Substitute values of $x$ around the vertex to determine key points for the parabola:

   • For $x = 0$: $f(0) = 2(0 - 1)^2 + 5 = 2(1) + 5 = 7 \quad \text{Point: } (0, 7).$
   • For $x = 2$ (symmetric to $x = 0$): $f(2) = 2(2 - 1)^2 + 5 = 2(1) + 5 = 7 \quad \text{Point: } (2, 7).$
   • For $x = -1$: $f(-1) = 2((-1) - 1)^2 + 5 = 2(4) + 5 = 13 \quad \text{Point: } (-1, 13).$
   • For $x = 3$ (symmetric to $x = -1$): $f(3) = 2(3 - 1)^2 + 5 = 2(4) + 5 = 13 \quad \text{Point: } (3, 13).$

4. Graph the points: Plot the vertex $(1, 5)$, points $(0, 7)$, $(2, 7)$, $(-1, 13)$, and $(3, 13)$. Then draw a smooth upward parabola.

Part (b)

Equation of the axis of symmetry: $x = 1$

Part (c)

Domain and range of the function:

1. Domain: Since $f(x)$ is a quadratic function, it is defined for all real values of $x$. Thus: $\text{Domain: } (-\infty, \infty).$

2. Range: The vertex $(1, 5)$ gives the minimum value of $f(x)$, and the parabola opens upwards. Hence: $\text{Range: } [5, \infty).$

Would you like a plotted graph or further clarification?

Here are 5 relative questions for deeper understanding:

1. How does the value of $a$ affect the steepness and direction of the parabola?
2. What would change in the graph if the vertex were shifted to $(-2, 3)$?
3. How can you determine the axis of symmetry directly from the vertex form?
4. What are the key differences between standard form and vertex form of a quadratic function?
5. How would the graph change if $a = -2$ instead of $2$?

Tip: Use symmetry when graphing quadratic functions—it simplifies calculations and ensures accuracy.