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The problem provided is about a function \( f: \mathbb{R} \to \mathbb{R} \), where \( f(x) = x^2 + 2x - 3 \). There are three parts to the problem:

1. **Part a:** Compute the function for specific values of \( x \), i.e., \( f(-4) \), \( f(-3) \), \( f(-2) \), \( f(-1) \), \( f(0) \), and \( f(2) \).
2. **Part b:** Plot the graph of the function.
3. **Part c:** Given the domain \( D_f = \{x \mid -4 \leq x \leq 2, x \in \mathbb{R}\} \), determine the range of the function over this domain.

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### Part a: Calculate \( f(x) \) for the given values of \( x \)

The function is \( f(x) = x^2 + 2x - 3 \). Let's compute:

- \( f(-4) = (-4)^2 + 2(-4) - 3 = 16 - 8 - 3 = 5 \)
- \( f(-3) = (-3)^2 + 2(-3) - 3 = 9 - 6 - 3 = 0 \)
- \( f(-2) = (-2)^2 + 2(-2) - 3 = 4 - 4 - 3 = -3 \)
- \( f(-1) = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4 \)
- \( f(0) = (0)^2 + 2(0) - 3 = -3 \)
- \( f(2) = (2)^2 + 2(2) - 3 = 4 + 4 - 3 = 5 \)

### Part b: Graph of the function
The function is a quadratic equation \( f(x) = x^2 + 2x - 3 \), which forms a parabola. The vertex of the parabola can be found using the formula for the vertex of a quadratic function, \( x_{\text{vertex}} = -\frac{b}{2a} \), where \( a = 1 \) and \( b = 2 \).

- \( x_{\text{vertex}} = -\frac{2}{2(1)} = -1 \)
- Substituting \( x = -1 \) into the function: \( f(-1) = -4 \).

Thus, the vertex of the parabola is \( (-1, -4) \), and the parabola opens upwards (since the coefficient of \( x^2 \) is positive). You can sketch the parabola based on this information, noting the symmetry around \( x = -1 \).

### Part c: Range of the function
The domain is given as \( D_f = \{x \mid -4 \leq x \leq 2\} \). To find the range, we evaluate the function at the boundaries of the domain and at the vertex (since it's a quadratic function):

- We already computed \( f(-4) = 5 \), \( f(-1) = -4 \), and \( f(2) = 5 \).
- The minimum value of \( f(x) \) occurs at the vertex, so the range of the function is \( \{y \mid -4 \leq y \leq 5\} \).

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### Summary of Answers:
1. **\( f(-4) = 5 \), \( f(-3) = 0 \), \( f(-2) = -3 \), \( f(-1) = -4 \), \( f(0) = -3 \), \( f(2) = 5 \)**
2. **The graph is a parabola with a vertex at \( (-1, -4) \) and it opens upwards.**
3. **The range is \( \{y \mid -4 \leq y \leq 5\} \).**

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Would you like further details or clarifications?

Here are 5 additional related questions:
1. How do you find the axis of symmetry for a quadratic function?
2. How can you determine if a parabola opens upwards or downwards?
3. What is the significance of the vertex in the graph of a quadratic function?
4. How do you find the roots of a quadratic equation like \( f(x) = x^2 + 2x - 3 \)?
5. How does the domain affect the range in quadratic functions?

**Tip:** For any quadratic function \( f(x) = ax^2 + bx + c \), the vertex represents the maximum or minimum point, depending on whether \( a \) is negative (maximum) or positive (minimum).

Diketahui fungsi f: R -> R dan f(x) = x^2 + 2x - 3. Hitunglah f(-4), f(-3), f(-2), f(-1), f(0), dan f(2). Gambarkan grafik fungsi tersebut! Jika daerah asal fungsi tersebut adalah D_f = {x | -4 <= x <= 2, x ∈ R}, tentukan daerah hasilnya!

Solve the quadratic function f(x) = x^2 + 2x - 3 by calculating f(-4), f(-3), f(-2), f(-1), f(0), and f(2). Graph the function and determine the range for the domain D_f = {-4 ≤ x ≤ 2}. This problem explores key algebraic concepts, including graphing and analyzing quadratic functions, suitable for students in Grades 9-11.

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