Problem 1 - Precalculus Mathematics for Calculus

Question

The solutions of the equation \(x^{2}-2 x-3=0\) are the ________ -intercepts of the graph of \(y=x^{2}-2 x-3\).

Step-by-Step Solution

Answer

The x-intercepts are 3 and -1.

Step 1: Identify the Quadratic Function

The given equation is a quadratic equation: \(x^2 - 2x - 3 = 0\). This can also be viewed as a quadratic function \(y = x^2 - 2x - 3\). The solutions to this equation represent the \(x\)-intercepts of the graph of the function, where the graph of \(y = x^2 - 2x - 3\) crosses the \(x\)-axis.

Step 2: Factor the Quadratic Expression

To find the \(x\)-intercepts, we need to solve \(x^2 - 2x - 3 = 0\). Start by factoring the quadratic expression. We need two numbers whose product is \(-3\) and whose sum is \(-2\). These numbers are \(-3\) and \(1\), so we can factor as \((x - 3)(x + 1) = 0\).

Step 3: Solve for x

Set each factor equal to zero to find the solutions: - \(x - 3 = 0 \Rightarrow x = 3\)- \(x + 1 = 0 \Rightarrow x = -1\).These values are the solutions to the equation and represent the \(x\)-intercepts.

Step 4: Conclusion on the x-Intercepts

The solutions \(x = 3\) and \(x = -1\) indicate that the \(x\)-intercepts of the graph of \(y = x^2 - 2x - 3\) are \((3, 0)\) and \((-1, 0)\).

Key Concepts

Factoring

Factoring is a common method for solving quadratic equations, especially when the expression can be neatly broken down into two binomial factors. In quadratic form, an equation typically looks like this: \[x^2 + bx + c = 0\] Our goal in factoring is to express this equation as a product of two simpler expressions. To do this, we need two numbers that multiply to the constant term (c) and add up to the linear coefficient (b). Let's consider the quadratic equation from the exercise: \[x^2 - 2x - 3 = 0\] Here, we’re searching for two numbers that multiply to \(-3\) (the constant term in our equation) and add to \(-2\) (the coefficient of x). These numbers are \(-3\) and \(1\). Now we can factor this equation as: \[(x - 3)(x + 1) = 0\] Factoring allows us to break down the expression easily so that we can find the solution to the quadratic equation by setting each factor equal to zero.

X-intercepts

The x-intercepts are the points where the graph of a function crosses the x-axis. For a quadratic function, these are precisely the solutions to the equation when the y-value is zero. This is because the x-axis represents the line where \(y = 0\). In the equation \(y = x^2 - 2x - 3\), finding the x-intercepts means solving \(x^2 - 2x - 3 = 0\). The solutions provide us with the x-coordinates of the points where the graph meets the x-axis.

• To find these intercepts, we can use methods such as factoring, completing the square, or the quadratic formula.
• In our case, the equation \((x - 3)(x + 1) = 0\) shows that the graph crosses the x-axis at points \((3, 0)\) and \((-1, 0)\).

Understanding x-intercepts helps in sketching the graph and knowing where the function changes signs.

Solving Quadratic Equations

Solving quadratic equations is about finding the values of x that make the equation equal to zero. There are several methods to achieve this, and choosing the right one often depends on the form the equation is presented in as well as the complexity of the coefficients involved.Here's a quick tour of popular methods:

• Factoring: As used in our exercise, when a quadratic is factorable, we can express it as a product of linear terms.
• Quadratic Formula: This powerful formula applies to all quadratic equations: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Ideal when factoring is complicated or not possible.
• Completing the Square: Involves manipulating the equation into a perfect square trinomial to find the roots.

In essence, each method seeks to uncover the same truth—the values of x where the quadratic expression meets itself at zero value. In practice, for the equation \(x^2 - 2x - 3 = 0\), using factoring was straightforward and yielded reliable results through the equations \(x-3 = 0\) and \(x+1 = 0\). By solving these, \(x = 3\) and \(x = -1\) instantly give us the sought solutions.