Problem 21
Question
In Exercises 19-22, graphically estimate the \( x \)- and \( y \)-intercepts of the graph. Verify your results algebraically. \( y = |x + 2| \)
Step-by-Step Solution
Verified Answer
The x-intercept of the function \(y = |x + 2|\) is -2 and the y-intercept is 2. These results were verified both graphically and algebraically.
1Step 1: Graphical representation
Draw the graph of the function \(y = |x + 2|\). The graph displays a V-shaped curve. The pivot point of the curve (also known as the vertex) is the point where the function crosses the y-axis. This point estimates the y-intercept. Also, note where the graph meets the x-axis, this is your estimate for the x-intercept.
2Step 2: Graphical estimation of intercepts
From the graph, it can be estimated that the curve intercepts the x-axis at x = -2, which is the x-intercept. Then, we can see that the curve meets the y-axis when the function value is positive. Hence, the y-intercept is 2.
3Step 3: Algebraic verification of x-intercept
To calculate the x-intercept algebraically, substitute \(y = 0\) in the function and solve for \(x\). So, \(0 = |x + 2|\). Which implies x = -2 when the modulus operator is removed. Thus, x-intercept is -2.
4Step 4: Algebraic verification of y-intercept
To calculate the y-intercept algebraically, substitute \(x = 0\) in the function, so \(y = |0 + 2|\), which results in \(y = 2\). Thus, y-intercept is 2.
Key Concepts
Graphical EstimationAbsolute Value FunctionsIntercepts of a Function
Graphical Estimation
Graphical estimation is a handy approach to predict values of certain points on a graph without performing detailed calculations. When looking at the graph of a function, like the V-shaped curve of the absolute value function \( y = |x + 2| \), you can estimate where the graph crosses the x-axis and the y-axis—these are known as intercepts.
To make a graphical estimation:
To make a graphical estimation:
- First, identify where the graph intersects the x-axis. This point is typically where the graph outputs zero (for the function \( y \)).
- Second, look at where the graph crosses the y-axis. This is found by determining the point where the input value \( x \) is zero.
Absolute Value Functions
Absolute value functions are a fundamental concept in algebra, often resulting in a distinctive V-shaped graph. An absolute value function, such as \( y = |x + 2| \), includes the absolute value operator which serves to make any input positive by converting negatives into positives.
The key characteristics of absolute value functions include:
The key characteristics of absolute value functions include:
- **Vertex**: The 'corner' or pivot of the V-shaped graph, marking a point of symmetry. For \( y = |x + 2| \), this vertex is at (-2, 0).
- **Non-Negative Output**: Since the absolute value cannot be negative, the graph stays entirely in the upper half of a coordinate grid, unless it's been shifted downwards.
- **Symmetry**: The graph is symmetrical around a vertical line passing through the vertex.
Intercepts of a Function
Intercepts in a function's graph provide valuable information about its behavior at specific points. For any function, intercepts refer to the points where the graph crosses the axes.
There are two main types of intercepts:
There are two main types of intercepts:
- **x-intercept**: This is the point where the graph crosses the x-axis. Here, the function's output (\( y \)) is zero, which allows us to solve for \( x \) by setting the equation to zero. For \( y = |x + 2| \), solving \( 0 = |x + 2| \) leads to two possible solutions due to the absolute value, but results in a single x-intercept at \( x = -2 \).
- **y-intercept**: This occurs where the graph meets the y-axis, and is found by substituting \( x = 0 \) into the function. For the function \( y = |x + 2| \), substituting gives \( y = 2 \), making the y-intercept at the point (0, 2).
Other exercises in this chapter
Problem 21
In Exercises 19-36, determine whether the equation represents \(y\) as a function of \(x\). \(x^2 + y = 4\)
View solution Problem 21
In Exercises 17-28, find the slope and \(y\)-intercept (if possible) of the equation of the line. Sketch the line. \( 5x - 2 = 0 \)
View solution Problem 21
In Exercise 15-24, determine the quadrant(s) in which \( (x, y) \) is located so that the condition(s) is (are) satisfied. \( x 0 \)
View solution Problem 22
In Exercises 17-28, evaluate the indicated function for \(f(x) = x^2 + 1\) and \(g(x) = x - 4\). \((f + g)(t-2)\)
View solution