Finding The Turning Point Of Quadratic Equation Y = 2x^2 + 5x - 2

The turning point of a quadratic equation's graph, also known as the vertex, represents a crucial feature of the parabola. It's the point where the curve changes direction, either from decreasing to increasing (a minimum point) or from increasing to decreasing (a maximum point). Finding the coordinates of this turning point is a common task in algebra and calculus, offering insights into the behavior of quadratic functions. This article dives deep into determining the coordinates of the turning point for the quadratic equation $y = 2x^2 + 5x - 2$. We'll explore different methods, providing a comprehensive understanding for students and enthusiasts alike. Understanding the turning point is not just an academic exercise; it has practical applications in various fields, such as optimization problems in engineering, economics, and physics.

Methods to Find the Turning Point

There are several methods to determine the turning point of a quadratic equation. We'll focus on two primary approaches: completing the square and using the vertex formula. Each method provides a unique perspective and reinforces key concepts in algebra. By mastering these techniques, you'll gain a deeper understanding of quadratic functions and their graphical representations.

1. Completing the Square

Completing the square is a powerful algebraic technique used to rewrite a quadratic equation in vertex form. The vertex form directly reveals the coordinates of the turning point. This method involves manipulating the equation to create a perfect square trinomial, allowing us to express the quadratic in the form $y = a(x - h)^2 + k$, where $(h, k)$ represents the coordinates of the vertex. For our equation, $y = 2x^2 + 5x - 2$, we'll walk through each step to illustrate the process.

First, factor out the coefficient of the $x^2$ term (which is 2 in this case) from the first two terms:

$y = 2(x^2 + \frac{5}{2}x) - 2$

Next, we need to add and subtract a value inside the parenthesis that will complete the square. To find this value, take half of the coefficient of the $x$ term (which is $\frac{5}{2}$), square it, and that's the value we'll add and subtract. Half of $\frac{5}{2}$ is $\frac{5}{4}$, and squaring it gives us $(\frac{5}{4})^2 = \frac{25}{16}$. So we add and subtract $\frac{25}{16}$ inside the parenthesis:

$y = 2(x^2 + \frac{5}{2}x + \frac{25}{16} - \frac{25}{16}) - 2$

Now, we can rewrite the expression inside the parenthesis as a perfect square trinomial:

$y = 2((x + \frac{5}{4})^2 - \frac{25}{16}) - 2$

Distribute the 2 to both terms inside the parenthesis:

$y = 2(x + \frac{5}{4})^2 - 2(\frac{25}{16}) - 2$

Simplify the expression:

$y = 2(x + \frac{5}{4})^2 - \frac{25}{8} - 2$

Convert 2 to a fraction with a denominator of 8:

$y = 2(x + \frac{5}{4})^2 - \frac{25}{8} - \frac{16}{8}$

Combine the constant terms:

$y = 2(x + \frac{5}{4})^2 - \frac{41}{8}$

Now the equation is in vertex form, $y = a(x - h)^2 + k$. Comparing this with our equation, we can identify the coordinates of the vertex as $(h, k) = (-\frac{5}{4}, -\frac{41}{8})$. Thus, the turning point of the graph is $(-\frac{5}{4}, -\frac{41}{8})$. This method provides a clear algebraic pathway to the solution, emphasizing the structure of quadratic equations and the significance of vertex form.

2. Using the Vertex Formula

An alternative and often quicker method to find the turning point is by using the vertex formula. This formula directly provides the x-coordinate of the vertex, which can then be substituted back into the original equation to find the y-coordinate. For a quadratic equation in the standard form $y = ax^2 + bx + c$, the x-coordinate of the vertex, denoted as $h$, is given by the formula:

$h = -\frac{b}{2a}$

For our equation, $y = 2x^2 + 5x - 2$, we have $a = 2$ and $b = 5$. Plugging these values into the formula, we get:

$h = -\frac{5}{2(2)} = -\frac{5}{4}$

This gives us the x-coordinate of the turning point. To find the y-coordinate, denoted as $k$, we substitute $x = -\frac{5}{4}$ back into the original equation:

$k = 2(-\frac{5}{4})^2 + 5(-\frac{5}{4}) - 2$

Now, we simplify the expression:

$k = 2(\frac{25}{16}) - \frac{25}{4} - 2$

$k = \frac{25}{8} - \frac{25}{4} - 2$

To combine these terms, we need a common denominator, which is 8. Convert the fractions accordingly:

$k = \frac{25}{8} - \frac{50}{8} - \frac{16}{8}$

Combine the numerators:

$k = \frac{25 - 50 - 16}{8} = \frac{-41}{8}$

Therefore, the y-coordinate of the turning point is $k = -\frac{41}{8}$. Combining the x and y coordinates, we find that the turning point is $(-\frac{5}{4}, -\frac{41}{8})$. This method provides a straightforward approach using a formula, making it particularly efficient for solving such problems. The vertex formula is a powerful tool that simplifies the process of finding the turning point, especially when dealing with more complex quadratic equations.

Comparing the Methods

Both completing the square and using the vertex formula are effective methods for finding the turning point of a quadratic equation. Completing the square provides a more in-depth understanding of the algebraic manipulation involved and leads to the vertex form of the equation, which is useful for various other analyses. On the other hand, the vertex formula offers a more direct and quicker route to the solution, especially when the focus is solely on finding the turning point's coordinates. Choosing the method depends on the specific context and the desired level of understanding. For instance, in situations requiring a transformation to vertex form, completing the square is the preferred choice. However, for quick calculations, the vertex formula is often more efficient.

The Significance of the Turning Point

The turning point, or vertex, of a quadratic equation's graph holds significant importance. It represents the minimum or maximum value of the quadratic function. If the coefficient of the $x^2$ term (i.e., $a$) is positive, the parabola opens upwards, and the turning point is the minimum. Conversely, if $a$ is negative, the parabola opens downwards, and the turning point is the maximum. This information is crucial in various applications, such as optimization problems. For instance, in business, the turning point can represent the point of maximum profit or minimum cost. In physics, it might represent the highest point reached by a projectile. Understanding the significance of the turning point allows us to interpret and apply quadratic functions in real-world scenarios.

Practical Applications

The turning point of a quadratic function has numerous practical applications across various disciplines. In physics, it helps determine the maximum height of a projectile's trajectory. In engineering, it can be used to optimize the design of structures, such as bridges and arches. In economics, it can represent the point of maximum profit or minimum cost. In mathematics, understanding the turning point is crucial for solving optimization problems and analyzing the behavior of quadratic functions. For instance, consider a scenario where a company wants to maximize its revenue. By modeling the revenue as a quadratic function, the turning point will represent the production level that yields maximum revenue. Similarly, in a physics context, if we model the height of a ball thrown in the air as a quadratic function, the turning point will give us the maximum height the ball reaches. These applications highlight the versatility and importance of understanding the turning point of quadratic equations.

Common Mistakes to Avoid

When finding the turning point of a quadratic equation, several common mistakes can occur. One frequent error is incorrectly applying the vertex formula, particularly mixing up the signs or miscalculating the fractions. Another common mistake is making errors during the process of completing the square, such as forgetting to factor out the coefficient of the $x^2$ term or incorrectly adding and subtracting the term needed to complete the square. Additionally, students sometimes struggle with substituting the x-coordinate of the vertex back into the original equation to find the y-coordinate, leading to incorrect y-values. To avoid these mistakes, it's crucial to practice each method step-by-step, double-checking each calculation, and understanding the underlying concepts. Regular practice and attention to detail are key to mastering these techniques and accurately finding the turning point of quadratic equations.

Conclusion

In conclusion, finding the turning point of the graph of the quadratic equation $y = 2x^2 + 5x - 2$ involves understanding the properties of quadratic functions and mastering the techniques of completing the square and using the vertex formula. By completing the square, we transformed the equation into vertex form, $y = 2(x + \frac{5}{4})^2 - \frac{41}{8}$, revealing the turning point as $(-\frac{5}{4}, -\frac{41}{8})$. Alternatively, using the vertex formula, $h = -\frac{b}{2a}$, we found the x-coordinate of the turning point to be $-\frac{5}{4}$, and substituting this back into the original equation gave us the y-coordinate of $-\frac{41}{8}$, confirming the turning point as $(-\frac{5}{4}, -\frac{41}{8})$. Both methods provide valuable insights and reinforce the significance of the turning point in various applications. This understanding is crucial for solving optimization problems and analyzing the behavior of quadratic functions in mathematics, physics, engineering, and economics. Remember to practice these methods diligently to avoid common mistakes and confidently apply them in different contexts. The turning point is not just a point on a graph; it's a key feature that unlocks a deeper understanding of quadratic functions and their real-world applications.

Therefore, the correct answer is:

B. $\left(-\frac{5}{4}, -\frac{41}{8}\right)$