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Chapter 3: Problem 33

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. $$f(x)=x^{2}+6 x+3$$

Short Answer

Expert verified
The vertex of the parabola is (-3,-6), the equation of the axis of symmetry is x=-3, the domain of the function is all real numbers, and the range of the function is [-6, 鈭�).

Step by step solution

01

- Find the vertex

First identify the vertex of the parabola. The x-coordinate of the vertex can be found using the formula \(x = -b/(2a)\). For the quadratic function \(f(x)=x^{2}+6 x+3\), b is 6 and a is 1. So, the x-coordinate of the vertex is \(-6/(2*1) = -3\). Now substitute x=-3 into the equation to find the y-coordinate of the vertex. \(f(-3)=(-3)^{2}+6*(-3)+3)=9-18+3=-6\). Hence, the vertex of the parabola is (-3,-6).
02

- Find the axis of symmetry

The axis of symmetry of a parabola is a vertical line that passes through the vertex. It can be expressed in the form \(x=k\), where k is the x-coordinate of the vertex. So, the equation of the axis of symmetry for this parabola is \(x=-3\).
03

- Determine the domain and range

The domain of the function is the set of all possible x-values. For a quadratic function, the domain is all real numbers, represented as (-鈭�, 鈭�). The range of the function is the set of possible y-values. Since the parabola opens upwards and the y-coordinate of the vertex is -6, the range of the function is [-6, 鈭�).
04

- Sketch the graph

Plot the points of the vertex (-3, -6). Draw the axis of symmetry (x=-3). Since the parabola opens upwards, the graph should mirror itself on the axis of symmetry. Finally, the downward vertex indicates that the graph opens up.

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