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Chapter 3: Problem 23

Use the vertex and intercepts to sketch the graph of each quadratic function. Give the equation of the parabola's axis of symmetry. Use the graph to determine the function's domain and range. $$f(x)=2(x+2)^{2}-1$$

Short Answer

Expert verified
The vertex is at \(-2, -1\). The axis of symmetry is the line \(x=-2\). The domain of the function is all real numbers while the range is \[-1, +\infty)\]

Step by step solution

01

Identify the Vertex

The standard form of a quadratic function is \(f(x)=a(x-h)^{2}+k\), where \((h,k)\) is the vertex of the parabola. In the given equation, \(f(x)=2(x+2)^{2}-1\), the vertex is at \(-2, -1\).
02

Finding the Axis of Symmetry

The axis of symmetry of a parabola is the vertical line \(x=h\), where \((h,k)\) is the vertex. For this exercise, the axis of symmetry is therefore the line \(x=-2\).
03

Sketch the Graph

The vertex \(-2, -1\) is the lowest point since the leading coefficient is positive, meaning the parabola opens upwards. Now, knowing the line of symmetry allows to sketch the parabola.
04

Identify the Domain and Range

For any quadratic function, the domain (permissible x values) is all real numbers (-infinity, infinity). Since the parabola opens upward and the vertex is its lowest point, the range (permissible y values) is \[-1, +\infty)\]

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