To analyze the quadratic function \( f(x) = 2(x + 1)^2 - 4 \), let's first identify the vertex, the y-intercept, and then sketch the graph.
Step 1: Identify the Vertex
The function is in vertex form \( f(x) = a(x - h)^2 + k \), where \( (h, k) \) is the vertex. From our function:
- \( a = 2 \)
- \( h = -1 \)
- \( k = -4 \)
Thus, the vertex is: \[ (-1, -4) \]
Step 2: Find the y-intercept
To find the y-intercept, set \( x = 0 \): \[ f(0) = 2(0 + 1)^2 - 4 = 2(1) - 4 = 2 - 4 = -2 \] So, the y-intercept is: \[ (0, -2) \]
Step 3: Axis of Symmetry
The axis of symmetry is given by the vertical line that passes through the vertex, so: \[ x = -1 \]
Step 4: Domain and Range
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Domain: The domain of a quadratic function is always all real numbers: \[ \text{Domain: } (-\infty, \infty) \]
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Range: Since the parabola opens upwards (because \( a = 2 > 0 \)), the range begins at the y-coordinate of the vertex and goes to infinity: \[ \text{Range: } [-4, \infty) \]
Summary Answers
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The axis of symmetry is: \[ x = -1 \]
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The domain of \( f \) is: \[ (-\infty, \infty) \]
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The range of \( f \) is: \[ [-4, \infty) \]
You can graph the function using the vertex at (-1, -4) and the y-intercept at (0, -2), along with plotting additional points if needed. The parabola will open upwards and will be symmetric about the line \( x = -1 \).