Problem 28 - Precalculus with Calculus Previews

Question

Find the \(x\) - and \(y\) -intercepts of the graph of the given function. Do not graph. $$ f(x)=4^{x} x^{4}-4^{x+1} $$

Step-by-Step Solution

Answer

The y-intercept is (0, -4), and the x-intercepts are (\(\sqrt{2}\), 0) and (-\(\sqrt{2}\), 0).

Step 1: Finding the y-intercept

The y-intercept of a function occurs where the graph intersects the y-axis. This happens when \(x = 0\). To find the y-intercept for \(f(x) = 4^x x^4 - 4^{x + 1}\), substitute \(x = 0\) into the function:\[f(0) = 4^0 \, 0^4 - 4^{0 + 1}\]Calculate:\[f(0) = 1 \, \cdot 0 - 4 = -4\]So, the y-intercept is \((0, -4)\).

Step 2: Finding the x-intercepts

The x-intercepts of a function occur where the graph intersects the x-axis. This is where \(f(x) = 0\). Thus, set the function equal to zero and solve for \(x\):\[4^x x^4 - 4^{x+1} = 0\]Factor the expression:\[4^x (x^4 - 4) = 0\]Here, either \(4^x = 0\) or \(x^4 - 4 = 0\):- \(4^x\) can never be zero as any positive number raised to any power remains positive.- Solve \(x^4 - 4 = 0\) using the difference of squares:\[(x^2 - 2)(x^2 + 2) = 0\]Solve each factor:- \(x^2 - 2 = 0\) gives \(x = \pm \sqrt{2}\) - \(x^2 + 2 = 0\) gives no real solutions since \(x^2 = -2\) is not possible for real numbers.Thus, the x-intercepts are \((\sqrt{2}, 0)\) and \((-\sqrt{2}, 0)\).

Key Concepts

Understanding Function Graphs

A function graph visually represents the relationship between the input values, often denoted as \( x \), and the output values, represented as \( f(x) \) or \( y \). To effectively understand and analyze these graphs, one must be familiar with key features such as intercepts, asymptotes, and the overall shape. The intercepts, both x- and y-intercepts, are points where the graph crosses the axes.

• Y-intercept: This is the point where the graph crosses the y-axis. It occurs when \( x = 0 \). To find it, substitute \( x = 0 \) in the function. In our problem, this yields the y-intercept as \((0, -4)\).
• X-intercepts: These are the points where the graph crosses the x-axis, indicating where the function value, \( f(x) \), equals zero. Solving this involves setting the function to zero and solving for \( x \).

Understanding function graphs helps in visualizing algebraic solutions and conveys intuitive understanding of numerical relationships. They are essential tools in solving real-world problems, providing insights into how functions behave across different values of \( x \).

Solving Equations for Intercepts

Solving equations is a fundamental skill in mathematics, especially when finding intercepts of a function graph. For any function \( f(x) \), intercepts are determined by solving specific equations:

• Finding the Y-intercept: Substitute \( x = 0 \) into \( f(x) \) to solve for \( y \). The solution gives the y-intercept value.
• Finding the X-intercepts: Set \( f(x) = 0 \) and solve this equation to find all possible \( x \) values where the function crosses the x-axis.

In our specific problem, solving \( 4^x x^4 - 4^{x + 1} = 0 \) involves isolating terms and simplifying using algebraic identities like the difference of squares. By simplifying and factoring, the equation \( x^4 - 4 = 0 \) is analyzed using well-known methods to find the x-intercepts \((\sqrt{2}, 0)\) and \((-\sqrt{2}, 0)\).

Application of the Difference of Squares

The difference of squares is a helpful algebraic identity used to simplify and solve polynomial equations. It applies to expressions of the form \( a^2 - b^2 \), and it is factored as \((a - b)(a + b)\).
In our problem, \( x^4 - 4 \) can be seen as a difference of squares, \( (x^2)^2 - 2^2 \), allowing us to factor it as \((x^2 - 2)(x^2 + 2)\).

• This identity aids in quickly identifying solutions for \( x \), simplifying the task of finding intercepts.
• It also highlights the importance of recognizing patterns and applying algebraic techniques to reduce complex problems into simpler ones.

Once factored, the equation \( (x^2 - 2)(x^2 + 2) = 0 \), when set to zero, simplifies the process of solving for \( x \). This is instrumental in finding the real solutions \( x = \pm \sqrt{2} \), while acknowledging \( x^2 + 2 = 0 \) yields no real solutions, since no real number squared equals a negative value.