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Chapter 3: Problem 19

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. $$f(x)=(x-1)^{2}+2$$

Short Answer

Expert verified
The vertex is at (1,2), and the equation of the axis of symmetry is x=1. The function intercepts the x-axis at (1,0). The domain of the function is all real values, and the range is all \(f(x) \geq 2\).

Step by step solution

01

Finding the Vertex

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

Determining the Axis of Symmetry

The axis of symmetry of a parabola passes through the vertex. The vertex is at x=1, so the equation of the axis of symmetry is x=1.
03

Finding x-Intercepts

The x-intercepts of the graph can be found by setting f(x) equal to 0. \(0=(x-1)^{2}+2\). Subtract 2 from both sides and take square root to get x=1 (this parabola only touches the x-axis at one point).
04

Sketching the Graph

Based on the above information, the graph can now be sketched. It's a parabola with the vertex at (1,2), and the axis of symmetry at x=1.
05

Determining the Domain and Range

For all quadratic functions, the domain is all real numbers. Since this graph opens upwards and the vertex is at (1,2), the range is \(f(x) \geq 2\).

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