Finding The X-Intercepts Of F(x)=5|x+4|-9 A Step-by-Step Guide

Introduction: Understanding $x$-intercepts and Absolute Value Functions

In mathematics, finding the $x$-intercepts, also known as horizontal intercepts, of a function is a fundamental concept in understanding the behavior and characteristics of that function. The x-intercepts are the points where the graph of the function intersects the $x$-axis. At these points, the $y$-value (or the function value, $f(x)$) is equal to zero. In simpler terms, we are looking for the values of $x$ that make $f(x) = 0$. This is crucial in various applications, such as determining the roots of an equation, analyzing the stability of systems, and understanding the graphical representation of a function. The function we are going to analyze is $f(x) = 5|x+4| - 9$, which involves an absolute value. Absolute value functions introduce a piecewise nature to the function, as the absolute value of a number is its distance from zero, irrespective of the sign. This means we need to consider two cases: when the expression inside the absolute value is positive or zero, and when it is negative. Solving for the $x$-intercepts of an absolute value function requires a careful consideration of these cases, ensuring that we account for all possible solutions. Understanding the nature of absolute value functions is important. The absolute value of a number, denoted as $|x|$, is its distance from zero on the number line. It is always non-negative. For example, $|3| = 3$ and $|-3| = 3$. This property leads to absolute value functions having a V-shape, with the vertex at the point where the expression inside the absolute value is zero. In our given function, $f(x) = 5|x+4| - 9$, the absolute value expression is $|x+4|$. This means we need to consider what happens when $x+4$ is positive or zero and when it is negative. By finding the $x$-intercepts, we gain valuable insights into where the function crosses the $x$-axis, helping us sketch its graph and understand its behavior. This process involves setting $f(x)$ to zero and solving for $x$, taking into account the two possible cases arising from the absolute value. This thorough approach ensures that we identify all points where the function intersects the $x$-axis, providing a comprehensive understanding of its horizontal intercepts.

Step-by-Step Solution: Finding the $x$-intercepts of $f(x)=5|x+4|-9$

To find the $x$-intercepts of the function $f(x) = 5|x+4| - 9$, we need to solve the equation $f(x) = 0$. This means we need to find the values of $x$ that make the function equal to zero. The presence of the absolute value requires us to consider two separate cases, as the absolute value function behaves differently depending on the sign of its argument. The first step in finding the x-intercepts is to set the function equal to zero: $5|x+4| - 9 = 0$ Next, we isolate the absolute value term by adding 9 to both sides of the equation: $5|x+4| = 9$ Then, we divide both sides by 5 to get the absolute value term alone: $|x+4| = \frac9}{5}$ Now, we consider the two cases arising from the absolute value. **Case 1 The expression inside the absolute value is positive or zero.** This means $x+4$ is greater than or equal to zero. In this case, the absolute value of $x+4$ is simply $x+4$. So, we have: $x+4 = \frac{95}$ To solve for $x$, we subtract 4 from both sides $x = \frac{95} - 4$ To combine the terms, we need a common denominator. We rewrite 4 as $\frac{20}{5}$ $x = \frac{95} - \frac{20}{5}$ $x = \frac{9-20}{5}$ $x = \frac{-11}{5}$ So, one $x$-intercept is $x = -\frac{11}{5}$. **Case 2 The expression inside the absolute value is negative.** This means $x+4$ is less than zero. In this case, the absolute value of $x+4$ is $-(x+4)$. So, we have: $-(x+4) = \frac{95}$ We distribute the negative sign $-x - 4 = \frac{95}$ To solve for $x$, we first add 4 to both sides $-x = \frac{95} + 4$ Again, we rewrite 4 as $\frac{20}{5}$ $-x = \frac{95} + \frac{20}{5}$ $-x = \frac{29}{5}$ Now, we multiply both sides by -1 to solve for $x$ $x = -\frac{29{5}$ So, the other $x$-intercept is $x = -\frac{29}{5}$. Therefore, the x-intercepts of the function $f(x) = 5|x+4| - 9$ are $x = -\frac{11}{5}$ and $x = -\frac{29}{5}$. These are the points where the graph of the function crosses the $x$-axis.

Verification: Checking the $x$-intercepts

After finding the $x$-intercepts, it is important to verify the solutions to ensure their correctness. This step helps to catch any potential errors made during the solving process and confirms that the values we found indeed make the function equal to zero. We found the x-intercepts to be $x = -\frac{11}{5}$ and $x = -\frac{29}{5}$. To verify these, we will substitute each value back into the original function $f(x) = 5|x+4| - 9$ and check if the result is zero. Verification for $x = -\frac{11}{5}$: Substitute $x = -\frac{11}{5}$ into the function: $f\left(-\frac11}{5}\right) = 5\left|-\frac{11}{5} + 4\right| - 9$ First, we need to simplify the expression inside the absolute value. We rewrite 4 as $\frac{20}{5}$ $f\left(-\frac{115}\right) = 5\left|-\frac{11}{5} + \frac{20}{5}\right| - 9$ Now, we combine the fractions $f\left(-\frac{115}\right) = 5\left|\frac{-11+20}{5}\right| - 9$ $f\left(-\frac{11}{5}\right) = 5\left|\frac{9}{5}\right| - 9$ The absolute value of $\frac{9}{5}$ is $\frac{9}{5}$ $f\left(-\frac{115}\right) = 5\left(\frac{9}{5}\right) - 9$ Now, we multiply $f\left(-\frac{115}\right) = 9 - 9$ $f\left(-\frac{11}{5}\right) = 0$ So, $x = -\frac{11}{5}$ is indeed an $x$-intercept. **Verification for $x = -\frac{29}{5}$** Substitute $x = -\frac{295}$ into the function $f\left(-\frac{295}\right) = 5\left|-\frac{29}{5} + 4\right| - 9$ Again, we rewrite 4 as $\frac{20}{5}$ $f\left(-\frac{295}\right) = 5\left|-\frac{29}{5} + \frac{20}{5}\right| - 9$ Combine the fractions $f\left(-\frac{295}\right) = 5\left|\frac{-29+20}{5}\right| - 9$ $f\left(-\frac{29}{5}\right) = 5\left|\frac{-9}{5}\right| - 9$ The absolute value of $-\frac{9}{5}$ is $\frac{9}{5}$ $f\left(-\frac{295}\right) = 5\left(\frac{9}{5}\right) - 9$ Multiply $f\left(-\frac{29{5}\right) = 9 - 9$ $f\left(-\frac{29}{5}\right) = 0$ So, $x = -\frac{29}{5}$ is also an $x$-intercept. Both solutions have been verified to make the function equal to zero, confirming that they are the correct $x$-intercepts. This verification process is a crucial step in problem-solving, ensuring accuracy and building confidence in the results.

Graphical Interpretation: Visualizing the $x$-intercepts

A graphical interpretation can provide a deeper understanding of the $x$-intercepts and their significance in the context of the function. By visualizing the function and its intercepts, we can gain insights into its behavior and properties. The x-intercepts are the points where the graph of the function intersects the $x$-axis. For the function $f(x) = 5|x+4| - 9$, we found the $x$-intercepts to be $x = -\frac{11}{5}$ and $x = -\frac{29}{5}$. These correspond to the points $\left(-\frac{11}{5}, 0\right)$ and $\left(-\frac{29}{5}, 0\right)$ on the coordinate plane. To visualize this, we can sketch the graph of the function. The function is an absolute value function, which typically has a V-shape. The general form of an absolute value function is $f(x) = a|x-h| + k$, where $(h, k)$ is the vertex of the V-shape. In our case, $f(x) = 5|x+4| - 9$, we can rewrite it as $f(x) = 5|x-(-4)| - 9$. This tells us that the vertex of the V-shape is at the point $(-4, -9)$. The coefficient 5 in front of the absolute value indicates the steepness of the V-shape; a larger coefficient means a steeper V. The graph opens upwards because the coefficient is positive. Now, we can plot the vertex $(-4, -9)$ and the $x$-intercepts $\left(-\frac{11}{5}, 0\right)$ and $\left(-\frac{29}{5}, 0\right)$ on the coordinate plane. Note that $-\frac{11}{5} = -2.2$ and $-\frac{29}{5} = -5.8$. The $x$-intercepts are located on the $x$-axis, to the left of the origin. The graph of the function will be a V-shape with the vertex at $(-4, -9)$, and the two arms of the V will pass through the $x$-intercepts. The left arm of the V passes through $\left(-\frac{29}{5}, 0\right)$, and the right arm passes through $\left(-\frac{11}{5}, 0\right)$. This graphical representation confirms our algebraic solution. The points where the graph crosses the $x$-axis are indeed at $x = -\frac{11}{5}$ and $x = -\frac{29}{5}$. Visualizing the function helps to understand the relationship between the algebraic solution and the geometric representation. It also provides a quick way to check the correctness of the solution; if the calculated $x$-intercepts do not align with the graph, there might be an error in the calculations.

Conclusion: Significance of $x$-intercepts

In conclusion, we have successfully found the $x$-intercepts of the function $f(x) = 5|x+4| - 9$. The $x$-intercepts, also known as horizontal intercepts, are the points where the graph of the function intersects the $x$-axis. These points are crucial in understanding the behavior and characteristics of the function. We systematically solved for the x-intercepts by setting the function equal to zero and considering the two cases that arise due to the absolute value. This led us to the solutions $x = -\frac{11}{5}$ and $x = -\frac{29}{5}$. We verified these solutions by substituting them back into the original function, confirming that they indeed make the function equal to zero. This verification step is essential in ensuring the accuracy of our results. Furthermore, we discussed the graphical interpretation of the $x$-intercepts. Visualizing the function and its intercepts provides a deeper understanding of their significance. The graph of the function is a V-shape, with the vertex at $(-4, -9)$, and the two arms of the V intersect the $x$-axis at the $x$-intercepts. Understanding $x$-intercepts is important for several reasons. They represent the points where the function's value is zero, which can have significant practical implications depending on the context of the problem. For example, in a real-world scenario where $f(x)$ represents profit, the $x$-intercepts would represent the break-even points, where the profit is zero. In physics, if $f(x)$ represents the height of a projectile, the $x$-intercepts would represent the points where the projectile hits the ground. Moreover, $x$-intercepts are crucial in sketching the graph of a function. Knowing the $x$-intercepts, along with other key features such as the vertex and any asymptotes, allows us to create an accurate representation of the function's behavior. In summary, finding the $x$-intercepts is a fundamental skill in mathematics. It involves algebraic manipulation, attention to detail, and a solid understanding of the properties of functions. The $x$-intercepts provide valuable information about the function, both in terms of its algebraic representation and its graphical behavior. By mastering this skill, we can gain a deeper understanding of the mathematical concepts and their applications in various fields.