Problem 27 - College Algebra with Corequisite Support

Question

For the following exercises, find the intercepts of the functions. $$ f(x)=x^{4}-16 $$

Step-by-Step Solution

Answer

The y-intercept is (0, -16). The x-intercepts are (2, 0) and (-2, 0).

Step 1: Understand the Problem

The exercise asks us to find the intercepts of the function \( f(x) = x^4 - 16 \). The intercepts include both the x-intercepts and y-intercepts.

Step 2: Finding the Y-Intercept

For any function \( f(x) \), the y-intercept is the point where the graph crosses the y-axis. This occurs when \( x = 0 \). Substitute \( x = 0 \) into the function to find \( f(0) \).\[ f(0) = 0^4 - 16 = -16 \] Thus, the y-intercept is \( (0, -16) \).

Step 3: Finding the X-Intercepts

The x-intercepts occur where the graph crosses the x-axis, which means \( f(x) = 0 \). Set \( f(x) = 0 \) and solve for \( x \).\[ x^4 - 16 = 0 \]

Step 4: Solving the Equation

To solve \( x^4 - 16 = 0 \), recognize that it is a difference of squares. Apply the difference of squares formula: \( a^2 - b^2 = (a - b)(a + b) \) where \( a = x^2 \) and \( b = 4 \).\[ x^4 - 16 = (x^2 - 4)(x^2 + 4) = 0 \]

Step 5: Solve Each Factor

Now solve each factor for \( x \).\( x^2 - 4 = 0 \) and \( x^2 + 4 = 0 \).\[ x^2 - 4 = 0 \] gives \( x^2 = 4 \), thus \( x = 2 \) or \( x = -2 \).\[ x^2 + 4 = 0 \] gives \( x^2 = -4 \), which has no real solutions.

Step 6: Conclude the X-Intercepts

The real x-intercepts are the solutions when \( x^2 - 4 = 0 \). Therefore, the x-intercepts of the function are \( (2, 0) \) and \( (-2, 0) \). Anything else yields no real solution.

Key Concepts

Difference of Squares

The concept of the difference of squares is a fundamental algebraic technique that simplifies polynomial expressions. It's based on the identity \( a^2 - b^2 = (a - b)(a + b) \). This identity is invaluable as it transforms a quadratic difference into a product of two binomials.

In the exercise, we have the function \( f(x) = x^4 - 16 \). Notice that \( x^4 \) is essentially \( (x^2)^2 \) and 16 is \( 4^2 \). This means we can apply the difference of squares formula. By stating \( a = x^2 \) and \( b = 4 \), we get:
\[ x^4 - 16 = (x^2 - 4)(x^2 + 4) \]
Switching from a single expression to a product allows us to find the roots or solutions more easily. This classic maneuver reveals deeper insights into behaviors of polynomials and their zeroes, helping to identify points where graphs intersect axes.

X-Intercepts

X-intercepts are the points where the graph of a function crosses the x-axis. Expressed mathematically, these intercepts occur when \( f(x) = 0 \). Finding x-intercepts involves setting the function equal to zero and solving for \( x \).

For our function, \( f(x) = x^4 - 16 \), setting it to zero results in:\[ x^4 - 16 = 0 \]
This equation can be simplified using the difference of squares, giving us \((x^2 - 4)(x^2 + 4) = 0\). Solving \( x^2 - 4 = 0 \) yields:\[ x^2 = 4 \]

• \( x = 2 \)
• \( x = -2 \)

Hence, the x-intercepts are the points \( (2, 0) \) and \( (-2, 0) \). The factor \( x^2 + 4 = 0 \) lacks real solutions, as it implies taking the square root of a negative number, not possible within the real number set.

Y-Intercepts

The y-intercept of a function is where the graph touches the y-axis. This is found by evaluating \( f(x) \) at \( x = 0 \). Basically, we substitute zero for \( x \) in the function.

In our specific case, consider the function \( f(x) = x^4 - 16 \). Replacing \( x \) with 0, we have:
\[ f(0) = 0^4 - 16 = -16 \]
This tells us that the graph crosses the y-axis at the point \( (0, -16) \). Unlike x-intercepts, for which both values of \( x \) must be found, the y-intercept is always a single point. It indicates the initial condition or starting point of the function's graph as it moves away in either direction of the axes.