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Magazine Chapter 3: Problem 11
For the following exercises, find the \(x\) - and \(y\) -intercepts of the graphs of each function. $$ f(x)=-2|x+1|+6 $$
Short Answer
Expert verified
The \(x\)-intercepts are \((2, 0)\) and \((-4, 0)\); the \(y\)-intercept is \((0, 4)\).
Step by step solution
01
Identify the Function
The function given is \( f(x) = -2|x+1| + 6 \). This is an absolute value function with vertical translation and reflection adjustments.
02
Find the y-intercept
To find the \(y\)-intercept, substitute \(x = 0\) into the function. \[f(0) = -2|0+1| + 6 = -2|1| + 6 = -2(1) + 6 = -2 + 6 = 4\]Thus, the \(y\)-intercept is \((0, 4)\).
03
Find the x-intercepts
To find the \(x\)-intercepts, set the function equal to zero and solve for \(x\):\[-2|x+1| + 6 = 0\]\[|x+1| = \frac{6}{2} = 3\]This gives two equations:1. \(x+1 = 3\) leading to \(x = 2\)2. \(x+1 = -3\) leading to \(x = -4\)Thus, the \(x\)-intercepts are \((2, 0)\) and \((-4, 0)\).
04
Verify the Intercepts
Substitute the \(x\)-intercepts \((2,0)\) and \((-4,0)\) into the original function to verify:For \(x = 2\):\[f(2) = -2|2+1| + 6 = -2(3) + 6 = 0\]For \(x = -4\):\[f(-4) = -2|-4+1| + 6 = -2(3) + 6 = 0\]Both calculations confirm that the points are indeed \(x\)-intercepts.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Value Function
An absolute value function like the one given, \( f(x) = -2|x+1| + 6 \), is a type of piecewise function. In general, an absolute value function can be written in the form \( f(x) = a|x-h| + k \), where \( a \), \( h \), and \( k \) are constants. The absolute value, denoted by \(|x|\), measures the distance from zero on a number line without considering direction鈥攕o it is always non-negative.
The specific example \( f(x) = -2|x+1| + 6 \) involves some transformations of the basic absolute value function \( |x| \):
The specific example \( f(x) = -2|x+1| + 6 \) involves some transformations of the basic absolute value function \( |x| \):
- There is a vertical reflection because of the negative sign in front of the absolute value.
- The function shifts horizontally left by 1 unit due to the \(+1\) inside the absolute value.
- It shifts vertically up by 6 units because of the \(+6\) added to the function.
X-Intercepts
The \( x \)-intercepts of a function are the points where the graph intersects the \( x \)-axis. To find these, we set the function equal to zero鈥攂ecause at the \( x \)-intercepts, \( y = 0 \). For the function \( f(x) = -2|x+1| + 6 \), this requires setting and solving the equation \(-2|x+1| + 6 = 0\).
Solving the equation, we isolate the absolute value term:
Solving the equation, we isolate the absolute value term:
- \(|x+1| = 3\)
- When \( x+1 = 3 \), you solve it to find \( x = 2 \). This gives one intercept \((2,0)\).
- When \( x+1 = -3 \), you solve it to find \( x = -4 \). This gives another intercept \((-4,0)\).
Y-Intercept
The \( y \)-intercept is a specific point where the graph intersects the \( y \)-axis. Here, \( x \) is always zero. In the expression \( f(x) = -2|x+1| + 6 \), finding the \( y \)-intercept involves simply plugging in \( x = 0 \).
Here's what it looks like when you calculate it:
Here's what it looks like when you calculate it:
- \(f(0) = -2|0+1| + 6\)
- \(= -2(1) + 6\)
- \(= -2 + 6\)
- \(= 4\)
Function Graphs
Graphing functions offers a visual insight into how the function behaves across different domains. For absolute value functions, the graph usually forms a V-shape, depending on transformations affecting its position and orientation.
For the function \( f(x) = -2|x+1| + 6 \), several graphical characteristics can be highlighted:
For the function \( f(x) = -2|x+1| + 6 \), several graphical characteristics can be highlighted:
- Because of the \(-2\), the graph opens downward, reflecting over the \( x \)-axis.
- The horizonal shift of \( +1 \) moves the vertex of the graph left by one unit from the origin \( (0, 0) \) to \( (-1, 6) \).
- This vertex serves as the lowest point due to the reflection effect, and the graph increases both ways from this vertex point.
- The graph's interaction with the \( x \)-axis at points \((2,0)\) and \((-4,0)\) further illustrates the range over which the function decreases then increases.
- Finally, the \( y \)-intercept provides a vertical reference point at \((0, 4)\), confirming part of the graph as it ascends negatively through this \( y \)-value after the vertex.
