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Chapter 3: Problem 21

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. $$y-1=(x-3)^{2}$$

Short Answer

Expert verified
The vertex of the parabola is at (3,1), the axis of symmetry is the line \(x = 3\), and the y-intercept is at (0,10). In the drawn parabola, the domain is all real values of x, \(-\infty < x < \infty\), and the range for y values is \(y \geq 1\).

Step by step solution

01

Identify the vertex

The vertex form of a quadratic equation is \(y = a(x-h)^2 + k\), where (h,k) is the vertex of the parabola. So from the given equation \(y-1=(x-3)^{2}\), the vertex is at (3,1).
02

Determine the axis of symmetry

The axis of symmetry of a parabola is the vertical line that passes through the vertex. In this case, x=h. So, the equation of the axis of symmetry is \(x=3\).
03

Find the y-intercept

The y-intercept is found by setting x=0 in our equation. So, solving the equation for x=0 gives \(y-1=(0-3)^2\), or \(y=10\). Therefore, the y-intercept is (0,10).
04

Sketch the graph

Start by drawing a grid and marking the vertex at (3,1). Draw the axis of symmetry, a vertical line through the vertex. Plot the y-intercept at (0,10). The parabola opens upwards because a>0 (from the equation y=a(x-h)^2+k). Draw the branches of the parabola on either side of the axis of symmetry, passing them through the y-intercept.
05

Identify domain and range

For any quadratic function, the domain is all real numbers represented by \(-\infty < x < \infty\). The range is calculated from the vertex and the direction in which the parabola opens. Since our parabola opens upward and the vertex is at (3,1), the range of our function is \(y \geq 1\).

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