Problem 11
Question
For the following exercises, find the \(x\) - and \(y\) -intercepts of the graphs of each function. \(f(x)=-2|x+1|+6\)
Step-by-Step Solution
Verified Answer
The y-intercept is 4; the x-intercepts are 2 and -4.
1Step 1: Find the y-intercept
To find the y-intercept, set \(x=0\) in the function and solve for \(f(x)\).\[ f(0) = -2|0+1| + 6 \] Simplify the expression:\[ f(0) = -2(1) + 6 = -2 + 6 = 4 \] So, the y-intercept is 4.
2Step 2: Find the x-intercepts
To find the x-intercepts, set \(f(x)=0\) and solve for \(x\).\[ -2|x+1| + 6 = 0 \] Rearrange the equation:\[ -2|x+1| = -6 \] Divide both sides by \(-2\):\[ |x+1| = 3 \] This gives two equations to solve for \(x\):1. \(x+1 = 3\)2. \(x+1 = -3\)Solve the first equation:\[ x+1 = 3 \ x = 3 - 1 = 2 \]Solve the second equation:\[ x+1 = -3 \ x = -3 - 1 = -4 \]So, the x-intercepts are \(x = 2\) and \(x = -4\).
Key Concepts
Absolute Value FunctionsFinding InterceptsGraphical Analysis of Functions
Absolute Value Functions
Absolute value functions are a type of mathematical function that incorporate the absolute value of a variable within their equation. The absolute value of a number is its distance from zero on the number line, regardless of direction. This means that for any real number, the absolute value is always non-negative.
In the case of the function \(f(x) = -2|x+1|+6\), the term \(|x+1|\) takes the absolute value of \(x+1\), making whatever is inside the modulus sign non-negative. The presence of the absolute value is crucial because it affects how the graph of the function behaves. Whenever the value inside the modulus is positive or zero, the expression stays as it is. But if it is negative, the expression flips to become positive.
Absolute value functions typically create a "V" shaped graph. In this case, the graph is inverted because of the negative sign in front of \(-2|x+1|\), flipping the typical "V" shape upside down. The absolute value dictates that as \(x\) moves away from \(-1\) in either direction, the values of the function decrease, forming two downward sloping lines.
In the case of the function \(f(x) = -2|x+1|+6\), the term \(|x+1|\) takes the absolute value of \(x+1\), making whatever is inside the modulus sign non-negative. The presence of the absolute value is crucial because it affects how the graph of the function behaves. Whenever the value inside the modulus is positive or zero, the expression stays as it is. But if it is negative, the expression flips to become positive.
Absolute value functions typically create a "V" shaped graph. In this case, the graph is inverted because of the negative sign in front of \(-2|x+1|\), flipping the typical "V" shape upside down. The absolute value dictates that as \(x\) moves away from \(-1\) in either direction, the values of the function decrease, forming two downward sloping lines.
Finding Intercepts
Finding the intercepts of a function helps us understand where the graph intersects the axes. These points can be crucial for graphical analysis and provide insight into the function's behavior. We generally calculate the intercepts in two main categories:
- Y-intercept: This is the point where the graph crosses the y-axis. To find it, set \(x = 0\) and solve for \(f(x)\). In this problem, the y-intercept is found by substituting 0 for \(x\) in the function, giving us \(f(0) = 4\).
- X-intercepts: These are the points where the graph crosses the x-axis. To find these, set the function equal to zero and solve for \(x\). For the given function, setting \(f(x) = 0\) and solving the corresponding equation result in the x-intercepts at \(x = 2\) and \(x = -4\).
Graphical Analysis of Functions
Graphical analysis is a way to visually interpret functions by examining their graphs. It often begins with plotting the basic points such as intercepts, and proceeds with analyzing the function's shape, slope, and other important characteristics.
For the function \(f(x) = -2|x+1|+6\), interpreting the absolute value and the negative coefficient is essential. The graph of this function starts at the y-intercept \((0, 4)\) and forms a clear inverted "V" shape due to the negative multiplier \(-2\).
The x-intercepts \((2, 0)\) and \((-4, 0)\) are crucial as they indicate where the graph touches the x-axis. These points help divide the graph into intervals for further analysis. Between the x-intercepts, the function remains either above or below the zero line, which can be categorical information for deeper mathematical problems.
Employing graphical analysis, it's much easier to hypothesize or confirm behaviors such as growth, decline, or symmetry in functions, because it allows practitioners to visualize the whole domain at once. With the given function, the symmetry around \(x=-1\), where the absolute value equals zero, is evident, showcasing the nature of how absolute values impact a graph.
For the function \(f(x) = -2|x+1|+6\), interpreting the absolute value and the negative coefficient is essential. The graph of this function starts at the y-intercept \((0, 4)\) and forms a clear inverted "V" shape due to the negative multiplier \(-2\).
The x-intercepts \((2, 0)\) and \((-4, 0)\) are crucial as they indicate where the graph touches the x-axis. These points help divide the graph into intervals for further analysis. Between the x-intercepts, the function remains either above or below the zero line, which can be categorical information for deeper mathematical problems.
Employing graphical analysis, it's much easier to hypothesize or confirm behaviors such as growth, decline, or symmetry in functions, because it allows practitioners to visualize the whole domain at once. With the given function, the symmetry around \(x=-1\), where the absolute value equals zero, is evident, showcasing the nature of how absolute values impact a graph.
Other exercises in this chapter
Problem 10
For the following exercises, determine whether the relation represents \(y\) as a function of \(x\). \(x=y^{2}\)
View solution Problem 11
For the following exercises, find \(f^{-1}(x)\) for each function. \(f(x)=\frac{x}{x+2}\)
View solution Problem 11
For the following exercises, describe how the graph of the function is a transformation of the graph of the original function \(f\). \(y=f(x+43)\)
View solution Problem 11
For the following exercise, find the indicated function given \(f(x)=2 x^{2}+1\) and \(g(x)=3 x-5\). (a) \(f(g(2))\) (b) \(f(g(x))\) (c) \(g(f(x))\) (d) \((g \c
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