Problem 19 - College Algebra

Question

For the following exercises, find the \(x\) - or \(t\) -intercepts of the polynomial functions. $$ f(x)=2 x^{4}+6 x^{2}-8 $$

Step-by-Step Solution

Answer

The x-intercepts are \(x = 1\) and \(x = -1\).

Step 1: Understand the Intercepts

The intercepts of a function are the points where the graph of the function crosses the axis. In this case, we are finding the points where the graph crosses the x-axis, also known as the x-intercepts. This occurs where the value of the function is zero, i.e., where \( f(x) = 0 \).

Step 2: Set the Equation to Zero

To find the x-intercepts, set the polynomial function equal to zero: \(2x^4 + 6x^2 - 8 = 0\).

Step 3: Simplify the Equation

Notice that the equation \(2x^4 + 6x^2 - 8 = 0\) is a quadratic in disguise. Let \( y = x^2 \), then the equation becomes \( 2y^2 + 6y - 8 = 0 \).

Step 4: Use the Quadratic Formula

To solve \(2y^2 + 6y - 8 = 0\), apply the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = 6\), and \(c = -8\).

Step 5: Calculate the Discriminant

First, calculate the discriminant \( \Delta = b^2 - 4ac = 6^2 - 4(2)(-8) = 36 + 64 = 100\). Since the discriminant is positive, there are two real roots.

Step 6: Find the Roots for y

Using the quadratic formula, \(y = \frac{-6 \pm \sqrt{100}}{4}\), gives \(y = \frac{-6 + 10}{4}\) or \(y = \frac{-6 - 10}{4}\). Simplify to obtain \(y = 1\) and \(y = -4\).

Step 7: Substitute Back to Solve for x

Recall \(y = x^2\). Therefore, for \(y = 1\), \(x^2 = 1\) which gives solutions \(x = 1\) and \(x = -1\). For \(y = -4\), \(x^2 = -4\), which has no real solutions, because the square of a real number cannot be negative.

Step 8: List the X-Intercepts

The x-intercepts of the function are \(x = 1\) and \(x = -1\).

Key Concepts

x-intercepts

The x-intercepts of a polynomial function are the points where the graph meets the x-axis. These points represent the solutions of the equation when the function equals zero. To find these intercepts, you must set the function equal to zero and solve it. In our case, with the function \( f(x) = 2x^4 + 6x^2 - 8 \), we find the x-intercepts by solving \( 2x^4 + 6x^2 - 8 = 0 \).

This process means we are looking for x values that make the function output zero. At these intercepts, the graph crosses the x-axis. It is important because x-intercepts provide valuable insights into the roots of the polynomial and the behavior of the graph.

• Set the equation to zero: \( f(x) = 0 \).
• Solve the resulting equation to find the intercepts.
• In this example, the x-intercepts are found to be \( x = 1 \) and \( x = -1 \).

quadratic formula

The quadratic formula is a powerful tool for finding the roots of a quadratic equation, which is an equation of the form \( ax^2 + bx + c = 0 \). It is defined as:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

In our exercise, the polynomial \( 2x^4 + 6x^2 - 8 = 0 \) can be rewritten as a quadratic equation by substituting \( y = x^2 \), to become \( 2y^2 + 6y - 8 = 0 \). We then apply the quadratic formula to find the values of \( y \).

• Identify \( a = 2 \), \( b = 6 \), \( c = -8 \).
• Substitute these values into the quadratic formula.
• Calculate the results to find \( y \) values, which then help us determine the corresponding x values.

Understanding the quadratic formula is crucial since it provides the exact solutions for any quadratic equation, including those derived from polynomial manipulations.

discriminant

The discriminant is a component of the quadratic formula that provides insight into the nature of the roots of a quadratic equation. It is given by the expression under the square root in the quadratic formula: \( b^2 - 4ac \).

The value of the discriminant tells us about the roots:

• If it is positive, the equation has two distinct real roots.
• If it is zero, there is exactly one real root (a repeated root).
• If it is negative, there are no real roots, only complex roots.

In the given problem, we calculated the discriminant as \( 100 \) (since \( 6^2 - 4 \times 2 \times (-8) = 100 \)). This positive value indicates two real roots for the quadratic \( 2y^2 + 6y - 8 = 0 \), supporting the finding of two x-values \( x = 1 \) and \( x = -1 \).
Examining the discriminant not only tells us how many solutions exist but provides a quick way to assess the nature of these solutions.

real solutions

Real solutions are the set of solutions to an equation that are real numbers, as opposed to complex numbers. For quadratic equations or disguised quadratic equations like our problem, whether solutions exist and their nature depends significantly on the discriminant.

In our process:\( x^2 = 1 \) led to two real solutions \( x = 1 \) and \( x = -1 \), derived from the intersecting points on the x-axis. However, \( x^2 = -4 \) produced no real solutions, since a square cannot result in a negative number within the real number system.

• Recognize that the roots that aren't real will not correspond to x-intercepts.
• Only where the discriminant allows real solutions do we find x-intercepts at those points.
  This distinction ensures clarity in how real numbers differ from complex numbers in solutions, guiding us towards understanding graph intersections and their relevance.