Find The Vertex Of $f(x)=\frac{1}{2} X^2+3 X+\frac{3}{2}$

Hey there, math whizzes and parabola pals! Today, we're diving deep into the fascinating world of quadratic functions, and our main quest is to find the vertex of a particular function: $f(x)=\frac{1}{2} x^2+3 x+\frac{3}{2}$. You know, that special point on the graph where the parabola either reaches its highest or lowest point. It's like the command center of your parabola, guys! Whether you're sketching graphs for a project, solving optimization problems, or just trying to flex those math muscles, understanding how to pinpoint the vertex is a super valuable skill. We're going to break it down step-by-step, making sure it’s not just understandable but actually fun. So, buckle up, grab your graphing paper (or your favorite digital tool), and let's unravel the secrets of this parabola together. We’ll explore different methods, explain the why behind the how, and make sure you leave here feeling like a vertex-finding pro. Get ready to transform those confusing equations into clear, visual insights!

The Standard Form and Why It Matters

Alright, before we can find the vertex of $f(x)=\frac{1}{2} x^2+3 x+\frac{3}{2}$, it's crucial to chat about the standard form of a quadratic equation. Most of the time, you'll see quadratics written as $f(x) = ax^2 + bx + c$. This is the bedrock, the foundation upon which we build our understanding. In our specific function, $f(x)=\frac{1}{2} x^2+3 x+\frac{3}{2}$, we can easily identify our coefficients: $a = \frac{1}{2}$, $b = 3$, and $c = \frac{3}{2}$. Now, why is this standard form so important, you ask? Because it gives us direct clues about the parabola's behavior. The coefficient '$a$' tells us if the parabola opens upwards (if $a > 0$, like in our case where $a=\frac{1}{2}$) or downwards (if $a < 0$). It also hints at how wide or narrow the parabola is – a larger absolute value of '$a$' means a narrower graph. The coefficient '$b$' influences the parabola's position and slope, and '$c$' is simply the y-intercept, the point where the parabola crosses the y-axis (because when $x=0$, $f(0)=c$). So, just by glancing at the equation in standard form, we can already predict some key features of our parabola. This initial understanding is super helpful as we move towards finding that all-important vertex. It’s like getting a sneak peek at the main character’s personality before the big reveal! Keep these coefficients in mind, guys, because they'll be our trusty sidekicks throughout this vertex-finding adventure.

Method 1: The $-b/2a$ Magic Formula

Okay, team, let's get down to business with our first, and arguably most popular, method for finding the vertex. This technique relies on a neat little formula derived from the properties of parabolas: the x-coordinate of the vertex is given by $x = -b / 2a$. It's like a secret handshake for locating the horizontal position of our parabola's turning point. Remember our function $f(x)=\frac{1}{2} x^2+3 x+\frac{3}{2}$? We've already identified $a = \frac{1}{2}$ and $b = 3$. Now, let's plug these values into our formula:

$x = -b / 2a$ $x = -(3) / (2 \times \frac{1}{2})$ $x = -3 / 1$ $x = -3$

Boom! Just like that, we've found the x-coordinate of our vertex. This tells us that the vertex lies on the vertical line $x = -3$. But we're not done yet! A vertex is a point, and a point needs both an x and a y-coordinate. To find the y-coordinate, we simply plug this x-value back into our original function $f(x)$. It's like sending our x-coordinate on a mission to find its corresponding y-partner.

So, we calculate $f(-3)$:

$f(-3) = \frac{1}{2}(-3)^2 + 3(-3) + \frac{3}{2}$ $f(-3) = \frac{1}{2}(9) - 9 + \frac{3}{2}$ $f(-3) = \frac{9}{2} - 9 + \frac{3}{2}$

To make this easier, let's find a common denominator, which is 2:

$f(-3) = \frac{9}{2} - \frac{18}{2} + \frac{3}{2}$ $f(-3) = \frac{9 - 18 + 3}{2}$ $f(-3) = \frac{-6}{2}$ $f(-3) = -3$

And there we have it! The y-coordinate of our vertex is -3. So, the vertex of the function $f(x)=\frac{1}{2} x^2+3 x+\frac{3}{2}$ is the point (-3, -3). This $-b/2a$ formula is a real lifesaver, guys, and it's incredibly efficient once you get the hang of it. It's a fundamental tool in our quadratic toolkit!

Method 2: Completing the Square – The Art of Transformation

Now, let's explore another powerful way to find the vertex: completing the square. This method is super cool because it not only reveals the vertex but also transforms our quadratic function into its vertex form, $f(x) = a(x-h)^2 + k$. In this form, $(h, k)$ is directly the vertex! It's like giving our function a makeover that instantly shows off its most important feature. For our function $f(x)=\frac{1}{2} x^2+3 x+\frac{3}{2}$, let's get started.

First, we want to isolate the terms with $x$ and factor out the coefficient of $x^2$ (which is '$a$'). In our case, $a=\frac{1}{2}$.

$f(x) = \frac{1}{2}(x^2 + 6x) + \frac{3}{2}$

Notice how we divided the $3x$ term by $\frac{1}{2}$ to get $6x$. Now, the magic of completing the square happens inside the parentheses. We want to turn the expression $x^2 + 6x$ into a perfect square trinomial. To do this, we take the coefficient of the $x$ term (which is 6), divide it by 2, and then square the result. So, $(6/2)^2 = 3^2 = 9$. We add this 9 inside the parentheses.

$f(x) = \frac{1}{2}(x^2 + 6x + 9) + \frac{3}{2}$

But here's the catch, guys: we didn't just add 9. We added $\frac{1}{2} \times 9$ to the entire function because of the factor outside the parentheses. So, we've effectively added $\frac{9}{2}$. To keep our equation balanced, we must subtract this same amount from the outside.

$f(x) = \frac{1}{2}(x^2 + 6x + 9) + \frac{3}{2} - \frac{9}{2}$

Now, the expression inside the parentheses is a perfect square: $(x+3)^2$. And we can simplify the constants on the outside.

$f(x) = \frac{1}{2}(x+3)^2 + \frac{3 - 9}{2}$ $f(x) = \frac{1}{2}(x+3)^2 + \frac{-6}{2}$ $f(x) = \frac{1}{2}(x+3)^2 - 3$

Voila! We've transformed our function into the vertex form $f(x) = a(x-h)^2 + k$. By comparing $f(x) = \frac{1}{2}(x+3)^2 - 3$ with the general vertex form, we can see that $a = \frac{1}{2}$, $h = -3$ (because it's $x - (-3)$), and $k = -3$. Therefore, the vertex $(h, k)$ is (-3, -3). This method is super illuminating because it gives you the vertex form directly, which is incredibly useful for graphing and understanding transformations. It’s a bit more involved than the $-b/2a$ formula, but the payoff in terms of insight is huge!

Method 3: Calculus – The Derivative's Insight

For those of you who have ventured into the realm of calculus, there's an even slicker way to find the vertex: using derivatives! This method leverages the fact that the vertex of a parabola is where the slope of the tangent line is zero. In simpler terms, it's the point where the graph momentarily flattens out before changing direction. We find this by taking the derivative of our function, setting it equal to zero, and solving for $x$. Let's give it a go with $f(x)=\frac{1}{2} x^2+3 x+\frac{3}{2}$.

First, we find the derivative, $f'(x)$. Using the power rule (bring the exponent down and multiply, then reduce the exponent by one), we get:

$f'(x) = \frac{d}{dx} (\frac{1}{2} x^2 + 3x + \frac{3}{2})$ $f'(x) = (\frac{1}{2} \times 2)x^{2-1} + (3 \times 1)x^{1-1} + 0$ $f'(x) = 1x^1 + 3x^0$ $f'(x) = x + 3$

Now, to find the x-coordinate of the vertex, we set this derivative equal to zero and solve for $x$:

$f'(x) = 0$ $x + 3 = 0$ $x = -3$

And just like that, we've found the x-coordinate of the vertex is -3! This matches what we got with the other methods, which is always a good sign. The derivative essentially tells us the slope of the function at any given point. At the vertex, the slope is zero, indicating a horizontal tangent. Once we have the x-coordinate, we proceed just as before: plug $x=-3$ back into the original function $f(x)$ to find the corresponding y-coordinate.

$f(-3) = \frac{1}{2}(-3)^2 + 3(-3) + \frac{3}{2}$ $f(-3) = \frac{1}{2}(9) - 9 + \frac{3}{2}$ $f(-3) = \frac{9}{2} - \frac{18}{2} + \frac{3}{2}$ $f(-3) = \frac{9 - 18 + 3}{2}$ $f(-3) = \frac{-6}{2}$ $f(-3) = -3$

So, using calculus, we also arrive at the vertex (-3, -3). This method is particularly elegant if you're comfortable with differentiation. It highlights a fundamental connection between the geometry of a curve and its algebraic representation through derivatives. Pretty neat, right guys?

Understanding the Vertex's Significance

The vertex of a parabola isn't just some random coordinate pair; it's the cornerstone of the parabola's graph. For $f(x)=\frac{1}{2} x^2+3 x+\frac{3}{2}$, we found our vertex to be at $(-3, -3)$. Since the coefficient $a = \frac{1}{2}$ is positive, our parabola opens upwards, meaning this vertex represents the minimum point on the graph. It's the absolute lowest the function's value will ever go. If '$a$' had been negative, the vertex would have been the maximum point. This point is crucial for several reasons. Firstly, it defines the axis of symmetry, which is a vertical line passing through the vertex. In our case, the axis of symmetry is the line $x = -3$. This line divides the parabola into two mirror images, making it incredibly useful for sketching the graph accurately. Any point on the parabola to the right of $x=-3$ has a corresponding point at the same height to the left of $x=-3$. Secondly, the vertex is fundamental in optimization problems. Whether you're trying to find the maximum height a projectile reaches, the minimum cost to produce a certain number of items, or the maximum area that can be enclosed by a fence, quadratic functions are often used to model these scenarios, and the vertex provides the optimal solution (either maximum or minimum). Understanding the vertex helps us interpret these real-world applications directly from the mathematical model. It's the point where the function reaches its extreme value, making it a focal point for analysis and problem-solving. So, when you find that vertex, remember you're not just finding a point; you're finding the key to the parabola's behavior and its potential applications!

Conclusion: Vertex Mastery Achieved!

Alright team, we've successfully navigated the journey to find the vertex of the function $f(x)=\frac{1}{2} x^2+3 x+\frac{3}{2}$, and we’ve done it using three distinct, powerful methods: the straightforward $-b/2a$ formula, the insightful vertex form derived from completing the square, and the elegant calculus approach using derivatives. Each method, while different, led us to the same undeniable truth: the vertex of this parabola is located at the coordinates (-3, -3). Remember, this point signifies the minimum value of our upward-opening parabola, and the line $x=-3$ serves as its axis of symmetry. Mastering the vertex is more than just a mathematical exercise; it's about unlocking a deeper understanding of how quadratic functions behave, how to graph them effectively, and how to apply them to solve real-world problems. Whether you're tackling homework, preparing for exams, or just enjoy the elegance of mathematics, knowing how to find the vertex is a fundamental skill that will serve you well. So, keep practicing, keep exploring, and never hesitate to use these tools to conquer any parabola that comes your way. You guys have got this! Happy graphing!