Chapter 5: Problem 14
For the following exercises, find the \(x\) - or \(t\) -intercepts of the polynomial functions. $$ f(x)=x^{3}+6 x^{2}-7 x $$
Short Answer
Expert verified
The x-intercepts are \( x = 0 \), \( x = 1 \), and \( x = -7 \).
Step by step solution
01
Understand the problem
We need to find the x-intercepts of the polynomial function \( f(x) = x^{3} + 6x^{2} - 7x \). X-intercepts occur where \( f(x) = 0 \). So we must solve the equation \( x^{3} + 6x^{2} - 7x = 0 \) to find the values of \( x \).
02
Factor the polynomial
Notice that each term in \( x^{3} + 6x^{2} - 7x \) contains \( x \), so we can factor \( x \) out of the polynomial. This gives us \( x(x^{2} + 6x - 7) = 0 \).
03
Solve for x
Once factored, the equation \( x(x^{2} + 6x - 7) = 0 \) shows that a solution occurs either when \( x = 0 \) or when \( x^{2} + 6x - 7 = 0 \).
04
Solve the quadratic equation
To solve \( x^{2} + 6x - 7 = 0 \), we can use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 6 \), and \( c = -7 \). Calculate the discriminant: \( b^2 - 4ac = 6^2 - 4(1)(-7) = 36 + 28 = 64 \). The discriminant is 64, so there are two real solutions.
05
Apply the quadratic formula
Using the quadratic formula, \( x = \frac{-6 \pm \sqrt{64}}{2 \cdot 1} \). Simplifying gives two solutions: \( x = \frac{-6 + 8}{2} = 1 \) and \( x = \frac{-6 - 8}{2} = -7 \).
06
Compile the x-intercepts
The x-intercepts of the polynomial function are \( x = 0 \), \( x = 1 \), and \( x = -7 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
x-intercepts
To understand x-intercepts, think of them as points where the graph of a polynomial function crosses the x-axis. In simpler terms, it's where the output of the function is zero. For any function \( f(x) \), the x-intercepts occur where the function is set to zero, i.e., \( f(x) = 0 \). These points are crucial since they can provide insights into the behavior of polynomial functions and aid in graphing them. Here's a quick way to identify them:
- Set the polynomial equation equal to zero.
- Solve for \( x \) to find where the function value reaches zero.
factoring
Factoring is a powerful algebraic method used to simplify polynomial equations and find their solutions. When you factor a polynomial, you're breaking it down into simpler expressions (or factors) that multiply together to give the original polynomial.
In the context of finding x-intercepts, factoring helps identify solutions where the product is zero. For example, if \( ab = 0 \), either \( a = 0 \) or \( b = 0 \), or both, need to be true for the equation to hold.In our exercise, we factor the polynomial \( f(x) = x^3 + 6x^2 - 7x \) by noting that each term contains \( x \). Thus, we factor out \( x \) from every term:
In the context of finding x-intercepts, factoring helps identify solutions where the product is zero. For example, if \( ab = 0 \), either \( a = 0 \) or \( b = 0 \), or both, need to be true for the equation to hold.In our exercise, we factor the polynomial \( f(x) = x^3 + 6x^2 - 7x \) by noting that each term contains \( x \). Thus, we factor out \( x \) from every term:
- The original polynomial becomes \( x(x^2 + 6x - 7) = 0 \).
- Solving for \( x \) now involves solving \( x = 0 \) or \( x^2 + 6x - 7 = 0 \).
quadratic formula
The quadratic formula is an essential tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It gives us the exact roots of the equation, even when the quadratic does not factor easily. The formula is expressed as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula derives from completing the square and is particularly useful for complicated quadratics.
Steps to use the quadratic formula:
Steps to use the quadratic formula:
- Identify the coefficients \( a \), \( b \), and \( c \) from the quadratic equation.
- Calculate the discriminant \( b^2 - 4ac \).
- Insert values into the formula and simplify.
discriminant in quadratic equations
The discriminant is a component of the quadratic formula that helps predict the nature of the roots without actually solving the equation. It is represented as \( \Delta = b^2 - 4ac \).Here's what the discriminant reveals:
- If \( \Delta > 0 \), the quadratic has two distinct real roots.
- If \( \Delta = 0 \), there is exactly one real root (a repeated root).
- If \( \Delta < 0 \), the roots are complex and not real.