Math Problem Statement

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!

Solution

The problem provided is about a function f:RRf: \mathbb{R} \to \mathbb{R}, where f(x)=x2+2x3f(x) = x^2 + 2x - 3. There are three parts to the problem:

  1. Part a: Compute the function for specific values of xx, i.e., f(4)f(-4), f(3)f(-3), f(2)f(-2), f(1)f(-1), f(0)f(0), and f(2)f(2).
  2. Part b: Plot the graph of the function.
  3. Part c: Given the domain Df={x4x2,xR}D_f = \{x \mid -4 \leq x \leq 2, x \in \mathbb{R}\}, determine the range of the function over this domain.

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

The function is f(x)=x2+2x3f(x) = x^2 + 2x - 3. Let's compute:

  • f(4)=(4)2+2(4)3=1683=5f(-4) = (-4)^2 + 2(-4) - 3 = 16 - 8 - 3 = 5
  • f(3)=(3)2+2(3)3=963=0f(-3) = (-3)^2 + 2(-3) - 3 = 9 - 6 - 3 = 0
  • f(2)=(2)2+2(2)3=443=3f(-2) = (-2)^2 + 2(-2) - 3 = 4 - 4 - 3 = -3
  • f(1)=(1)2+2(1)3=123=4f(-1) = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4
  • f(0)=(0)2+2(0)3=3f(0) = (0)^2 + 2(0) - 3 = -3
  • f(2)=(2)2+2(2)3=4+43=5f(2) = (2)^2 + 2(2) - 3 = 4 + 4 - 3 = 5

Part b: Graph of the function

The function is a quadratic equation f(x)=x2+2x3f(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, xvertex=b2ax_{\text{vertex}} = -\frac{b}{2a}, where a=1a = 1 and b=2b = 2.

  • xvertex=22(1)=1x_{\text{vertex}} = -\frac{2}{2(1)} = -1
  • Substituting x=1x = -1 into the function: f(1)=4f(-1) = -4.

Thus, the vertex of the parabola is (1,4)(-1, -4), and the parabola opens upwards (since the coefficient of x2x^2 is positive). You can sketch the parabola based on this information, noting the symmetry around x=1x = -1.

Part c: Range of the function

The domain is given as Df={x4x2}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)=5f(-4) = 5, f(1)=4f(-1) = -4, and f(2)=5f(2) = 5.
  • The minimum value of f(x)f(x) occurs at the vertex, so the range of the function is {y4y5}\{y \mid -4 \leq y \leq 5\}.

Summary of Answers:

  1. f(4)=5f(-4) = 5, f(3)=0f(-3) = 0, f(2)=3f(-2) = -3, f(1)=4f(-1) = -4, f(0)=3f(0) = -3, f(2)=5f(2) = 5
  2. The graph is a parabola with a vertex at (1,4)(-1, -4) and it opens upwards.
  3. The range is {y4y5}\{y \mid -4 \leq y \leq 5\}.

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)=x2+2x3f(x) = x^2 + 2x - 3?
  5. How does the domain affect the range in quadratic functions?

Tip: For any quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c, the vertex represents the maximum or minimum point, depending on whether aa is negative (maximum) or positive (minimum).

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Graphing

Formulas

f(x) = ax^2 + bx + c
x_vertex = -b / 2a

Theorems

Quadratic Formula

Suitable Grade Level

Grades 9-11