Problem 30 - College Algebra

Question

Find the intercepts of the functions. $$f(x)=(x+3)\left(4 x^{2}-1\right)$$

Step-by-Step Solution

Answer

x-intercepts: -3, ±1/2; y-intercept: -3.

Step 1: Understand the Problem

The task is to find the x-intercepts and y-intercepts of the given function. An x-intercept occurs where the function equals zero (\(f(x) = 0\)), while the y-intercept occurs where \(x = 0\).

Step 2: Find the x-intercepts

To find the x-intercepts, set the function equal to zero: \[f(x) = (x+3)(4x^2-1) = 0.\]The equation is equal to zero when either \((x+3) = 0\) or \((4x^2-1) = 0\).Solving \(x+3 = 0\), we get \(x = -3\).For \(4x^2-1 = 0\), add 1 to both sides to obtain \(4x^2 = 1\), then divide by 4 to get \(x^2 = \frac{1}{4}\). Taking the square roots gives us \(x = \pm \frac{1}{2}\).Thus, the x-intercepts are \(x = -3\), \(x = \frac{1}{2}\), and \(x = -\frac{1}{2}\).

Step 3: Determine the y-intercept

To find the y-intercept, substitute \(x = 0\) into the function:\[f(0) = (0+3)(4(0)^2-1) = 3 \times (-1) = -3.\]Thus, the y-intercept is \(y = -3\).

Step 4: Summarize the Intercepts

The x-intercepts of the function are \(x = -3\), \(x = \frac{1}{2}\), and \(x = -\frac{1}{2}\). The y-intercept is \(y = -3\).

Key Concepts

x-intercepts

Finding x-intercepts is like finding out where a graph crosses the x-axis. This happens where the function itself equals zero. For our exercise, we set the function \( f(x) = (x+3)(4x^2-1) \) equal to zero. This means solving the equation \( (x+3)(4x^2-1) = 0 \).
When you have a product that equals zero, one or more of its factors must be zero:

• First, consider \( x+3 = 0 \). Solving this gives us \( x = -3 \).
• Next, for \( 4x^2-1 = 0 \), add 1 to both sides to get \( 4x^2 = 1 \). Divide by 4, resulting in \( x^2 = \frac{1}{4} \). Taking the square root, we find \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \).

Thus, the x-intercepts are at \( x = -3 \), \( x = \frac{1}{2} \), and \( x = -\frac{1}{2} \). These points tell us where the graph touches or crosses the x-axis.

y-intercepts

The y-intercept is the point where a graph crosses the y-axis, which means you set the x-value to zero and solve for the function's value. In our exercise, we substitute \( x = 0 \) into the function \( f(x) = (x+3)(4x^2-1) \).
Calculate:

• First, calculate \( 0+3 = 3 \).
• Next, compute \( 4(0)^2-1 = -1 \).
• Finally, multiply these results: \( 3 \times (-1) = -3 \).

So, the y-intercept is \( y = -3 \). This indicates where our graph cuts across the y-axis, specifically at the point \( (0,-3) \).

factoring polynomials

Factoring polynomials is a fundamental skill in algebra that involves expressing a polynomial as a product of its simpler factors. In our exercise, we saw \( f(x) = (x+3)(4x^2-1) \) already factored into two parts: \( (x+3) \) and \( (4x^2-1) \).
Why is factoring useful?

• It simplifies solving polynomial equations. By breaking down complex expressions into simpler parts, solving for x becomes straightforward.
• It helps in finding intercepts. As seen, finding x-intercepts is easier when dealing with products of factors.
• It aids in graphing polynomials. Knowing the factors allows you to see how the graph behaves at different points.

Always look for opportunities to factor polynomials, as it often simplifies many mathematical problems.

solving equations

Solving equations is about finding the value(s) of variables that make an equation true. It often involves several algebraic techniques.
In our exercise, we solved the equation \( (x+3)(4x^2-1) = 0 \). The strategy was to set each factor equal to zero individually:

• For \( x+3=0 \), simply subtract 3 from both sides to get \( x = -3 \).
• For \( 4x^2-1=0 \), we rearranged to find \( 4x^2 = 1 \), then \( x^2 = \frac{1}{4} \). Taking the square root, we found two solutions: \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \).

By understanding different methods for solving equations, such as factoring, isolating terms, and utilizing basic algebraic manipulations, we can find all potential solutions.