Problem 29 - A Graphical Approach to College Algebra

Question

It is not apparent from the standard viewing window whether the graph of the quadratic function intersects the \(x\) -axis once, twice, or not at all. Experiment with various windows to find the number of \(x\) -intercepts. If there are \(x\) -intercepts, give their coordinates to the nearest hundredth. (GRAPH CANT COPY) $$y=x^{2}-4.25 x+4.515$$

Step-by-Step Solution

Answer

The graph intersects the x-axis twice at (2.15, 0) and (2.10, 0).

Step 1: Understand the Equation

The given quadratic function is \( y = x^2 - 4.25x + 4.515 \). To determine the number of \( x \)-intercepts, we need to find the roots of this equation by setting \( y = 0 \) and solving for \( x \).

Step 2: Calculate the Discriminant

The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( b^2 - 4ac \). For our equation, \( a = 1 \), \( b = -4.25 \), and \( c = 4.515 \). Calculate the discriminant: \( (-4.25)^2 - 4 \times 1 \times 4.515 = 18.0625 - 18.06 = 0.0025 \).

Step 3: Interpret the Discriminant

Since the discriminant \( 0.0025 \) is greater than zero, the quadratic equation has two distinct real roots. This means the graph intersects the \( x \)-axis at two points.

Step 4: Find the Roots using the Quadratic Formula

The quadratic formula to find the roots is \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). Substituting the values: \( x = \frac{4.25 \pm \sqrt{0.0025}}{2} \). Calculate the square root of the discriminant: \( \sqrt{0.0025} = 0.05 \).

Step 5: Compute the \( x \)-intercepts

Substitute \( \sqrt{0.0025} = 0.05 \) into the formula: \( x_1 = \frac{4.25 + 0.05}{2} = 2.15 \) and \( x_2 = \frac{4.25 - 0.05}{2} = 2.10 \). Thus, the roots are \( x_1 \approx 2.15 \) and \( x_2 \approx 2.10 \).

Step 6: State the \( x \)-intercepts

The coordinates of the \( x \)-intercepts of the graph, accurate to the nearest hundredth, are approximately \((2.15, 0)\) and \((2.10, 0)\).

Key Concepts

x-intercepts

The concept of x-intercepts is crucial for understanding the positions where a graph intersects the x-axis. Think of x-intercepts as the points which solve the equation when the function value is zero. In our given quadratic function, it appears as: - **Definition**: The x-intercepts are the points where the function crosses the x-axis. These are the solutions to \( y = 0 \) in the equation - **Example Calculation**: For the function \( y = x^2 - 4.25x + 4.515 \), setting \( y = 0 \) allows us to find the x-intercepts.
Finding these intercepts requires understanding the relationship between the quadratic function and its graph. These points illustrate on a graph where the output of the quadratic function equals zero. Therefore, they provide critical insights into the nature and behavior of the quadratic equation.

discriminant

The discriminant acts as a tool to depict the nature of the roots of a quadratic equation. It simplistically informs us about the number and type of x-intercepts there may be when given a quadratic function. Here's how it works:- **Definition**: The discriminant is found using, \( b^2 - 4ac \), from the equation \( ax^2 + bx + c = 0 \). - **Determine the Roots**: - If the discriminant \( > 0 \), there are two real and distinct roots. - If it is \( = 0 \), you have one real double root. - If \( < 0 \), it results in no real roots.In our example, the discriminant was \( 0.0025 \), which is slightly greater than zero. Thus, our quadratic equation possesses two distinct real roots, confirming that the graph of this function will intersect the x-axis at two separate points.
Utilizing the discriminant simplifies predictions about graph behaviors without requiring extensive calculations, making it a quintessential concept in the analysis of quadratic functions.

quadratic formula

The quadratic formula provides a systematic method for finding the roots of any quadratic equation, even when they aren't readily factorable. By substituting directly into the formula, solutions can readily be determined:- **Formula Understanding**: The quadratic roots are given by: \[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\] - **Application**: In our case, by substituting \( a = 1 \), \( b = -4.25 \), and \( c = 4.515 \) into the formula, we found the roots as: \[x_1 = \frac{4.25 + 0.05}{2} = 2.15, \quad x_2 = \frac{4.25 - 0.05}{2} = 2.10\]
This formula serves as a robust tool for handling any quadratic equation, offering a clear path to solving for roots where other methods might prove cumbersome or ineffectual. It is especially valuable when manual factoring is difficult or impossible.