Problem 10 - College Algebra with Corequisite Support

Question

For the following exercises, find the \(x\) - and \(y\) -intercepts of the graphs of each function. \(f(x)=-3|x-2|-1\)

Step-by-Step Solution

Answer

There are no x-intercepts; the y-intercept is (0, -7).

Step 1: Identify the Equation of the Function

The given function is \( f(x) = -3|x-2| - 1 \). This is an absolute value function.

Step 2: Solve for x-intercept(s)

To find the x-intercept(s), set \( f(x) = 0 \) and solve for \( x \). This leads to the equation \(-3|x-2| - 1 = 0\). Add 1 to both sides to get \(-3|x-2| = 1\). Divide both sides by -3 to get \(|x-2| = -\frac{1}{3}\). Since \(|x-2|\) cannot be negative, this equation has no solution. Therefore, there is no x-intercept.

Step 3: Solve for y-intercept

To find the y-intercept, set \( x = 0 \) and solve for \( f(x) \). Substitute \( x = 0 \) into the function: \( f(0) = -3|0-2| - 1 \). Calculate \( f(0) = -3|2| - 1 = -3(2) - 1 = -6 - 1 = -7 \). Thus, the y-intercept is \( (0, -7) \).

Key Concepts

x-intercepts

The x-intercepts of a function are the points where the graph crosses the x-axis. At these points, the output of the function, or the y-value, is zero. To find the x-intercepts of a function, you need to set the function equal to zero and solve for x. This means you'll look for points where the equation becomes zero.In our example, we have the absolute value function:\[ f(x) = -3|x-2| - 1 \]To find the x-intercepts, we set it equal to zero:\[ -3|x-2| - 1 = 0 \]First, add 1 to both sides of the equation:\[ -3|x-2| = 1 \]Then, divide both sides by -3:\[ |x-2| = -\frac{1}{3} \]Here is where something important happens. The absolute value expression \(|x-2|\) cannot be negative, since absolute values are always zero or positive. In this case, \(-\frac{1}{3}\) is negative, which means no value of x can satisfy this equation. Consequently, this function has no x-intercepts.

y-intercepts

Finding the y-intercepts involves determining where the graph of a function crosses the y-axis. This happens when the input value, x, is equal to zero. For our given function:\[ f(x) = -3|x-2| - 1 \]Setting x to zero gives us:\[ f(0) = -3|0-2| - 1 \]Calculate the absolute value:\[ f(0) = -3|2| - 1 \]Simplify further:\[ f(0) = -3(2) - 1 = -6 - 1 = -7 \]Therefore, the y-intercept of the function is the point \( (0, -7) \). This point tells us that if you plot the function on graph paper, the graph will cross the y-axis at -7.

solving equations

When it comes to absolute value functions, solving equations often involves isolating the absolute value expression and then considering both positive and negative scenarios.The absolute value function: \[ f(x) = -3|x-2| - 1 \]does not require us to consider both cases because the equation \[ |x-2| = -\frac{1}{3} \]is invalid due to the negative on the right side. Why is this important? Absolute values measure distance and cannot output negative values. This understanding simplifies solving absolute value equations, especially when no solution is present because of a negative assignment to an absolute value. Understanding this can help prevent unnecessary complexity while solving absolute-value-based equations.

graphing functions

Graphing functions, especially those involving absolute values, requires understanding certain key properties.Absolute value functions typically form a 'V' shape. The graph of the function \[ f(x) = -3|x-2| - 1 \]will reflect this shape but it is inverted and shifted. - The \(-3\) factor in front of the absolute value indicates the graph is flipped vertically, meaning it opens downwards.- The \(-1\) indicates a vertical shift downwards by one unit, moving the vertex of the function lower on the y-axis.- Inside the absolute value function, \(x-2\) dictates the horizontal shift. Here the graph shifts 2 units to the right along the x-axis.These characteristics involve simple transformations. By following them, you can plot the function accurately. Typically, you identify and plot the vertex, which for this function is at \( (2, -1) \). From there, you can determine additional points by considering values of x close to 2 and applying the function to see how the graph progresses, maintaining the 'V' shape. This helps create a clear, accurate graph that visually represents the function.