Rewriting $y = 9x^2 + 9x - 1$ In Vertex Form A Comprehensive Guide

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Converting a quadratic equation from standard form to vertex form is a crucial skill in algebra, enabling us to easily identify the vertex of the parabola and understand its transformations. In this comprehensive guide, we will delve into the process of rewriting the given quadratic equation, $y = 9x^2 + 9x - 1$, in vertex form. We will explore the underlying concepts, step-by-step procedures, and potential pitfalls to ensure a thorough understanding. This exploration will empower you to confidently tackle similar problems and gain a deeper appreciation for the elegance of quadratic equations.

Understanding Vertex Form

Before we embark on the conversion process, let's first solidify our understanding of the vertex form of a quadratic equation. The vertex form is expressed as:

y=a(x−h)2+ky = a(x - h)^2 + k

where:

  • (a) represents the leading coefficient, which determines the parabola's direction and stretch.
  • (h, k) represents the vertex of the parabola, which is the point where the parabola changes direction. It's the minimum point if a > 0 and the maximum point if a < 0.

The vertex form provides a clear view of the parabola's vertex, making it easier to graph the equation and analyze its properties. The standard form of a quadratic equation, on the other hand, is given by:

y=ax2+bx+cy = ax^2 + bx + c

While the standard form is useful for identifying the y-intercept (which is c), it doesn't directly reveal the vertex. Therefore, converting from standard form to vertex form is a valuable technique.

Completing the Square: The Key to Conversion

The primary method for converting a quadratic equation from standard form to vertex form is completing the square. This technique involves manipulating the equation algebraically to create a perfect square trinomial, which can then be factored into a squared binomial. Let's break down the process step-by-step for the given equation, $y = 9x^2 + 9x - 1$.

Step 1: Factor out the leading coefficient from the $x^2$ and $x$ terms

Our first step is to factor out the leading coefficient, which is 9 in this case, from the terms containing $x^2$ and $x$:

y=9(x2+x)−1y = 9(x^2 + x) - 1

This step isolates the quadratic and linear terms, preparing them for the completion of the square.

Step 2: Complete the square inside the parentheses

Now, we focus on the expression inside the parentheses: $x^2 + x$. To complete the square, we need to add and subtract a specific value that will transform this expression into a perfect square trinomial. The value we need to add is the square of half the coefficient of the x term. In this case, the coefficient of the x term is 1, so half of it is $\frac{1}{2}$, and its square is $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$.

We add and subtract this value inside the parentheses:

y=9(x2+x+14−14)−1y = 9\left(x^2 + x + \frac{1}{4} - \frac{1}{4}\right) - 1

Notice that we are effectively adding 0, which doesn't change the equation's value. However, this manipulation allows us to create a perfect square trinomial.

Step 3: Rewrite the perfect square trinomial as a squared binomial

The expression $x^2 + x + \frac{1}{4}$ is now a perfect square trinomial, which can be factored as $\left(x + \frac{1}{2}\right)^2$. Substitute this back into the equation:

y=9((x+12)2−14)−1y = 9\left(\left(x + \frac{1}{2}\right)^2 - \frac{1}{4}\right) - 1

Step 4: Distribute the leading coefficient

Next, we distribute the 9 back into the parentheses:

y=9(x+12)2−9(14)−1y = 9\left(x + \frac{1}{2}\right)^2 - 9\left(\frac{1}{4}\right) - 1

This simplifies to:

y=9(x+12)2−94−1y = 9\left(x + \frac{1}{2}\right)^2 - \frac{9}{4} - 1

Step 5: Combine the constant terms

Finally, we combine the constant terms to obtain the vertex form:

y=9(x+12)2−94−44y = 9\left(x + \frac{1}{2}\right)^2 - \frac{9}{4} - \frac{4}{4}

y=9(x+12)2−134y = 9\left(x + \frac{1}{2}\right)^2 - \frac{13}{4}

Therefore, the vertex form of the quadratic equation $y = 9x^2 + 9x - 1$ is $y = 9\left(x + \frac{1}{2}\right)^2 - \frac{13}{4}$.

Identifying the Vertex

From the vertex form, $y = 9\left(x + \frac{1}{2}\right)^2 - \frac{13}{4}$, we can easily identify the vertex of the parabola. Recall that the vertex form is given by $y = a(x - h)^2 + k$, where (h, k) is the vertex.

In our case, we have:

  • h = -$\frac{1}{2}$ (note the subtraction in the general form)
  • k = -$\frac{13}{4}$

Thus, the vertex of the parabola is (-$\frac{1}{2}$, -$\frac{13}{4}$). Since the leading coefficient a is 9, which is positive, the parabola opens upwards, and the vertex represents the minimum point of the parabola.

Comparing with the Given Options

Now, let's compare our result with the options provided:

A. $y=9\left(x+\frac{1}{2}\right)^2-1$ B. $y=9\left(x+\frac{1}{2}\right)^2+\frac{5}{4}$ C. $y=9\left(x+\frac{1}{2}\right)^2-\frac{13}{4}$ D. Discussion category: mathematics

Our derived vertex form, $y = 9\left(x + \frac{1}{2}\right)^2 - \frac{13}{4}$, matches option C. Therefore, option C is the correct answer.

Common Mistakes and How to Avoid Them

Converting quadratic equations to vertex form can be tricky, and several common mistakes can occur. Let's discuss these pitfalls and how to avoid them:

  1. Forgetting to factor out the leading coefficient: A crucial first step is to factor out the leading coefficient from the $x^2$ and x terms. Failing to do so will lead to an incorrect completion of the square.

    • How to avoid: Always make factoring the leading coefficient your first step.

  2. Incorrectly calculating the value to complete the square: The value to add and subtract is the square of half the coefficient of the x term. Forgetting to halve the coefficient or squaring the wrong value will result in an incorrect vertex form.

    • How to avoid: Double-check your calculation: (coefficient of x / 2)^2.

  3. Forgetting to distribute the leading coefficient after completing the square: After rewriting the perfect square trinomial as a squared binomial, remember to distribute the leading coefficient back into the parentheses. Failing to do so will alter the equation's value.

    • How to avoid: Make it a habit to distribute the leading coefficient immediately after rewriting the trinomial.

  4. Sign errors: Pay close attention to signs, especially when dealing with negative numbers. A single sign error can lead to an incorrect vertex.

    • How to avoid: Be meticulous with your signs and double-check your work.

  5. Misinterpreting the vertex form: Remember that the vertex form is $y = a(x - h)^2 + k$, so the x-coordinate of the vertex is h, not -h. Forgetting this can lead to misidentifying the vertex.

    • How to avoid: Always refer back to the general form $y = a(x - h)^2 + k$ to ensure you're correctly extracting h and k.

By being aware of these common mistakes and practicing diligently, you can master the art of converting quadratic equations to vertex form.

Applications of Vertex Form

The vertex form of a quadratic equation is not just a mathematical curiosity; it has practical applications in various fields. Here are a few examples:

  1. Optimization problems: In many real-world scenarios, we need to find the maximum or minimum value of a quantity. Quadratic functions are often used to model these situations, and the vertex represents the maximum or minimum point. For instance, if we have a quadratic function representing the profit of a business, the vertex will tell us the production level that maximizes profit.

  2. Projectile motion: The path of a projectile, such as a ball thrown in the air, can be modeled by a quadratic function. The vertex of the parabola represents the maximum height reached by the projectile.

  3. Engineering design: Quadratic equations are used in designing various structures, such as bridges and arches. The vertex form helps engineers determine the optimal shape and dimensions of these structures.

  4. Curve fitting: In data analysis, we often need to find a mathematical function that best fits a set of data points. Quadratic functions are commonly used for curve fitting, and the vertex form can provide valuable insights into the shape and characteristics of the curve.

By understanding the vertex form, you can unlock the power of quadratic equations to solve real-world problems.

Conclusion

In this comprehensive guide, we have explored the process of rewriting the quadratic equation $y = 9x^2 + 9x - 1$ in vertex form. We have learned the importance of completing the square, identified common mistakes to avoid, and discussed the practical applications of vertex form. By mastering this technique, you will gain a deeper understanding of quadratic equations and their role in various fields. Remember to practice regularly, and don't hesitate to revisit this guide whenever you need a refresher. With dedication and perseverance, you can confidently conquer any quadratic equation challenge.