Problem 36
Question
Give the domain of the function and sketch the graph. $$f(x)=-\frac{1}{2} x-3$$
Step-by-Step Solution
Verified Answer
The domain of the function \(f(x) = -\frac{1}{2} x - 3\) is the set of all real numbers. The graph of the function is a straight line sloping downward to the right with a y-intercept at -3.
1Step 1: Identify the Domain
In general, the domain of any linear equation is the set of all real numbers, because the variables in a linear equation can be any real number. Thus, the domain of the function \(f(x) = -\frac{1}{2} x - 3\) is also the set of all real numbers.
2Step 2: Identify the Slope and Y-intercept
The slope of the given function \(f(x) = -\frac{1}{2} x - 3\) is the coefficient of \(x\), which is \(-\frac{1}{2}\). The y-intercept is the constant term, which is -3.
3Step 3: Sketch the Graph
Plotting the y-intercept at -3 on the y-axis forms the initial point. From there, the slope is to go down one unit (because the slope is -1/2) for every two units moved to the right. This will form a straight line tilted to the right.
Key Concepts
Domain of a FunctionGraphingSlope-Intercept FormReal Numbers
Domain of a Function
The domain of a function describes all the possible values that the input variable, often represented as \(x\), can take. In the case of a linear function like \(f(x) = -\frac{1}{2} x - 3\), the domain is all real numbers.
This means that any real number can be substituted for \(x\) in the function. There are no restrictions because the equation is defined everywhere on the number line.
This means that any real number can be substituted for \(x\) in the function. There are no restrictions because the equation is defined everywhere on the number line.
- For rational and radical functions, domains can be limited by division by zero or taking the square root of a negative number.
- However, for linear functions, including the given one, such problems do not exist.
Graphing
Graphing a linear function involves plotting points on a coordinate plane to form a straight line. The line represents all possible solutions to the equation.
For the function \(f(x) = -\frac{1}{2} x - 3\), start by identifying key elements like the y-intercept and slope. These will guide the plotting process.
For the function \(f(x) = -\frac{1}{2} x - 3\), start by identifying key elements like the y-intercept and slope. These will guide the plotting process.
- Begin at the y-intercept: This is the point where the line intersects the y-axis. For our function, it’s -3.
- Use the slope for direction: Here, the slope is \(-\frac{1}{2}\), meaning we move 1 unit down for every 2 units to the right.
Slope-Intercept Form
The slope-intercept form is an efficient way to express linear equations as \(y = mx + b\). In this structure, \(m\) denotes the slope, while \(b\) represents the y-intercept.
Our function, \(f(x) = -\frac{1}{2} x - 3\), fits this form perfectly:
Our function, \(f(x) = -\frac{1}{2} x - 3\), fits this form perfectly:
- The slope \(m\) is \(-\frac{1}{2}\), indicating the line’s steepness and direction.
- The y-intercept \(b\) is -3, showing the point at which the line crosses the y-axis.
Real Numbers
The set of real numbers includes all the numbers that can be found on the number line. This includes both rational and irrational numbers.
Rational numbers are those that can be expressed as a fraction, such as integers and terminating or repeating decimals. Irrational numbers, on the other hand, include numbers that cannot be written as simple fractions, like \(\pi\) and \(\sqrt{2}\).
Rational numbers are those that can be expressed as a fraction, such as integers and terminating or repeating decimals. Irrational numbers, on the other hand, include numbers that cannot be written as simple fractions, like \(\pi\) and \(\sqrt{2}\).
- Real numbers form the foundation of linear functions, as they cover all possible values for \(x\) in examples like \(f(x) = -\frac{1}{2} x - 3\).
- They ensure that the domain of such functions is comprehensive and unrestricted.
Other exercises in this chapter
Problem 36
Find \(f\) such that \(f \circ g=F\) given that $$g(x)=x^{2}, F(x)=a x^{2}+b$$
View solution Problem 36
Give the domain and range of the function. $$f(x)=\frac{1}{2} \sqrt{1-4 x^{2}}$$
View solution Problem 36
Find the number(s) \(x\) in the interval \([0.2 \pi]\) which satisfy the equation. $$\sin 2 x=-\sqrt{3} / 2$$.
View solution Problem 36
Sketch the set on a number line. (-4,3].
View solution