Converting $y=(x+3)^2+(x+4)^2$ To Vertex Form A Step-by-Step Guide

Jul 20, 2025 by ADMIN 67 views

Rewriting $y=(x+3)^2+(x+4)^2$ in Vertex Form

Hey guys! Today, we're diving into a super common algebra problem: rewriting a quadratic equation into vertex form. Specifically, we're going to tackle the equation $y=(x+3)^2+(x+4)^2$ and transform it into that sleek vertex form. Trust me, once you get the hang of this, it'll be a piece of cake! We will walk through step by step how to convert a quadratic equation from its standard form to vertex form. Understanding this process is crucial for identifying key features of parabolas, such as the vertex, axis of symmetry, and maximum or minimum values. So, let's get started and make sure you grasp every detail!

Understanding Vertex Form

Before we jump into the problem, let's quickly recap what vertex form actually is. Vertex form of a quadratic equation is expressed as:

$y = a(x - h)^2 + k$

Where:

$(h, k)$ represents the vertex of the parabola.
$a$ determines the direction the parabola opens (upwards if $a > 0$ , downwards if $a < 0$ ) and its stretch.

Knowing this form is super handy because the vertex, that crucial turning point of the parabola, is right there in the equation! It tells us a lot about the parabola's behavior and where it sits on the coordinate plane. To understand why vertex form is so useful, consider what it immediately tells us about the parabola. The vertex $(h, k)$ gives the minimum or maximum point of the function, depending on the sign of $a$ . If $a$ is positive, the parabola opens upwards, and the vertex is the minimum point. Conversely, if $a$ is negative, the parabola opens downwards, and the vertex is the maximum point. This information is invaluable in various applications, such as optimization problems where you need to find the maximum or minimum value of a quadratic function. Furthermore, the vertex form makes it easy to identify the axis of symmetry, which is the vertical line $x = h$ . This line divides the parabola into two symmetrical halves, simplifying graphing and analysis. By converting a quadratic equation to vertex form, we gain direct access to these critical features, making it easier to understand and work with the function. So, the vertex form is not just a different way of writing a quadratic equation; it’s a powerful tool for unlocking key information about the parabola it represents.

Let's Expand the Given Equation

Alright, let's get our hands dirty with the given equation: $y = (x + 3)^2 + (x + 4)^2$ . The first thing we need to do is expand those squared terms. Remember the formula $(a + b)^2 = a^2 + 2ab + b^2$ ? We'll use that here.

Expanding $(x + 3)^2$ gives us:

$x^2 + 2(x)(3) + 3^2 = x^2 + 6x + 9$

And expanding $(x + 4)^2$ gives us:

$x^2 + 2(x)(4) + 4^2 = x^2 + 8x + 16$

Now, let's plug these back into our original equation:

$y = (x^2 + 6x + 9) + (x^2 + 8x + 16)$

Next, we combine like terms to simplify. This involves adding the coefficients of terms with the same power of $x$ . We combine the $x^2$ terms, the $x$ terms, and the constant terms separately to make the equation more manageable.

Combining Like Terms

Now we need to tidy things up by combining like terms. We have $x^2$ terms, $x$ terms, and constant terms. Let's group them together:

$y = (x^2 + x^2) + (6x + 8x) + (9 + 16)$

Adding these up gives us:

$y = 2x^2 + 14x + 25$

Awesome! We've now got our equation in standard quadratic form: $y = ax^2 + bx + c$ , where $a = 2$ , $b = 14$ , and $c = 25$ . But we're not done yet – we need to get it into vertex form. This step is crucial because the standard form, while useful for some purposes, doesn't immediately reveal the vertex of the parabola. The vertex form, as we discussed earlier, makes the vertex readily apparent, which is why we're undertaking this transformation. To get there, we'll use a technique called completing the square, which will allow us to rewrite the quadratic expression in a way that highlights the vertex. So, let's move on to the next step and transform this standard form into the coveted vertex form.

Completing the Square

Okay, here comes the fun part: completing the square! This is the key to transforming our equation into vertex form. The goal here is to rewrite the quadratic expression as a squared term plus a constant. This process might seem a bit tricky at first, but with practice, it becomes second nature.

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

$y = 2(x^2 + 7x) + 25$

Now, we need to figure out what to add inside the parentheses to make it a perfect square trinomial. Remember, a perfect square trinomial can be factored into the form $(x + n)^2$ or $(x - n)^2$ . To find this magic number, we take half of the coefficient of the $x$ term (which is 7), square it, and add it inside the parentheses. Half of 7 is $rac{7}{2}$ , and squaring it gives us $rac{49}{4}$ .

But hold on! We can't just add $rac{49}{4}$ inside the parentheses without balancing the equation. Since we're actually adding $2 imes rac{49}{4}$ (because of the 2 outside the parentheses), we need to subtract the same amount outside the parentheses to keep the equation balanced.

So, we have:

$y = 2igl(x^2 + 7x + rac{49}{4}igr) + 25 - 2 rac{49}{4}$

Now, the expression inside the parentheses is a perfect square trinomial, and we can rewrite it as a squared term.

Rewriting as Vertex Form

Now we can rewrite the expression inside the parentheses as a squared term. Remember, we added $rac{49}{4}$ inside the parentheses, which is the square of half the coefficient of our $x$ term (which was 7). So, the perfect square trinomial factors beautifully into:

$x^2 + 7x + rac{49}{4} = igl(x + rac{7}{2}igr)^2$

Let's substitute that back into our equation:

$y = 2igl(x + rac{7}{2}igr)^2 + 25 - 2 rac{49}{4}$

Now, let's simplify the constant term. We have $25 - 2 rac{49}{4}$ , which simplifies to:

$25 - rac{49}{2} = rac{50}{2} - rac{49}{2} = rac{1}{2}$

So, our equation becomes:

$y = 2igl(x + rac{7}{2}igr)^2 + rac{1}{2}$

And there you have it! We've successfully rewritten the equation in vertex form. This is a significant achievement because we can now easily identify the vertex of the parabola and understand its orientation.

Identifying the Vertex

Look at that! We've got our equation in vertex form: $y = 2igl(x + rac{7}{2}igr)^2 + rac{1}{2}$ . Now, let's pinpoint the vertex. Remember the vertex form equation: $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex.

In our equation, we can see that:

$a = 2$
$h = - rac{7}{2}$ (notice the minus sign in the formula!)
$k = rac{1}{2}$

So, the vertex of our parabola is $igl(- rac{7}{2}, rac{1}{2}igr)$ . That's super useful information! We now know the lowest point (since $a$ is positive) of our parabola. The vertex is a critical point for understanding the behavior of the quadratic function. It represents the minimum or maximum value of the function and serves as a reference point for graphing. Knowing the vertex allows us to quickly sketch the parabola and determine its key characteristics.

Furthermore, the vertex helps us identify the axis of symmetry, which is the vertical line that passes through the vertex. In this case, the axis of symmetry is $x = - rac{7}{2}$ . The axis of symmetry divides the parabola into two symmetrical halves, making it easier to analyze and graph the function. By finding the vertex, we gain valuable insights into the parabola's shape and position in the coordinate plane. This information is essential for solving various problems involving quadratic functions, such as optimization problems, projectile motion, and curve fitting.

Choosing the Correct Option

Now that we've transformed the equation and found the vertex form, let's go back to the original question and choose the correct option. We found that the equation $y = (x + 3)^2 + (x + 4)^2$ in vertex form is:

$y = 2igl(x + rac{7}{2}igr)^2 + rac{1}{2}$

Looking at the options provided, we can see that option B matches our result perfectly:

B. $y=2igl(x+ rac{7}{2}igr)^2+ rac{1}{2}$

So, B is the correct answer!

Key Takeaways

Alright, guys, let's recap what we've learned today. We took a quadratic equation, expanded it, combined like terms, and then used the awesome technique of completing the square to rewrite it in vertex form. We then identified the vertex of the parabola. That's a lot! Remember, the key steps are:

Expand: Expand any squared terms.
Combine: Combine like terms to get the standard form $y = ax^2 + bx + c$ .
Factor: Factor out the coefficient of the $x^2$ term from the first two terms.
Complete the Square: Add and subtract the square of half the coefficient of the $x$ term inside the parentheses.
Rewrite: Rewrite the trinomial as a squared term.
Simplify: Simplify the equation to get the vertex form $y = a(x - h)^2 + k$ .
Identify the vertex: The vertex is $(h, k)$ .

Understanding how to convert quadratic equations to vertex form is a fundamental skill in algebra. It not only helps in graphing parabolas but also in solving various real-world problems involving quadratic functions. The vertex form provides a clear picture of the parabola's turning point, which is crucial in optimization problems, physics, and engineering. By mastering this technique, you'll be well-equipped to tackle more complex mathematical challenges. So, keep practicing, and you'll become a pro at completing the square and rewriting equations in vertex form!

Practice Makes Perfect

The best way to really nail this down is to practice! Try rewriting other quadratic equations in vertex form. You can even make up your own equations and see if you can do it. The more you practice, the more comfortable you'll become with the process. Remember, each step builds upon the previous one, so understanding the underlying logic is crucial. Don't just memorize the steps; try to understand why each step is necessary to achieve the final result. This deeper understanding will make it easier to apply the technique to different problems and variations.

Also, don't hesitate to use online resources or textbooks for additional examples and explanations. Many websites and educational platforms offer interactive tools and step-by-step solutions that can help you visualize the process and check your work. Collaboration with classmates or study groups can also be beneficial. Discussing the concepts and working through problems together can clarify any doubts and reinforce your understanding. By actively engaging with the material and seeking out opportunities to practice, you'll build confidence and proficiency in rewriting quadratic equations in vertex form. So, keep up the good work, and you'll soon master this valuable skill!

Keep up the great work, and you'll be a vertex form whiz in no time! Remember, math is a journey, not a destination. Enjoy the process of learning, and don't be afraid to make mistakes – they're just opportunities to learn and grow! Happy solving!