The problem involves analyzing a quadratic function $f(x) = (x - 1)^2 + 5$ and identifying important features like the axis of symmetry, domain, and range. Let's break it down step by step.

Step 1: Axis of Symmetry

The function is written in vertex form: $f(x) = (x - h)^2 + k$ Here, $h = 1$ and $k = 5$. The axis of symmetry is a vertical line passing through $x = h$. Thus: $\text{Axis of symmetry: } x = 1$

Step 2: Domain

Since the quadratic function is defined for all real numbers, the domain is: $\text{Domain: } (-\infty, \infty)$

Step 3: Range

The vertex of the parabola is at $(1, 5)$. Since the parabola opens upwards (because the coefficient of $(x-1)^2$ is positive), the function's values are greater than or equal to $k = 5$. Thus: $\text{Range: } [5, \infty)$

Step 4: Graphing

1. Vertex: $(1, 5)$.
2. Y-Intercept: Substitute $x = 0$ into the function: $f(0) = (0 - 1)^2 + 5 = 1 + 5 = 6$ So, the y-intercept is $(0, 6)$.
3. Other points: Pick symmetric points around $x = 1$. For example:
   • $x = 2$: $f(2) = (2 - 1)^2 + 5 = 6$.
   • $x = 3$: $f(3) = (3 - 1)^2 + 5 = 9$.
   • Similarly for $x = 0, -1$, etc.

Final Details

• Plot the vertex, axis of symmetry, and points like $(0, 6)$, $(2, 6)$, $(3, 9)$, and $(-1, 9)$.
• Ensure the graph is symmetric about $x = 1$.

Would you like a sketch of the graph or further clarification?

5 Related Questions

1. How do we determine whether a parabola opens upward or downward?
2. Why does the axis of symmetry pass through the vertex?
3. How does the coefficient of $x^2$ affect the shape of the parabola?
4. What happens to the range if the vertex form has $-k$ instead of $+k$?
5. How do you find the x-intercepts of a parabola like this one?

Tip

Always start graphing a quadratic function by identifying its vertex and axis of symmetry—it simplifies plotting significantly!