Quadratic Function Vertex & X-Intercepts Explained

Hey there, math enthusiasts! Welcome back to Plastik Magazine, your go-to spot for all things numbers and equations. Today, we're diving deep into the fascinating world of quadratic functions. You know, those nifty parabolic shapes that pop up everywhere from projectile motion to designing rollercoasters. Specifically, we're going to tackle a common challenge: finding the vertex and x-intercepts of a quadratic function. This skill is super fundamental in understanding the behavior and key features of these curves. We'll be working with the function $f(x)=9.8 x^2+11.2 x+3.2$, and don't worry if you're not a calculus whiz; we'll break it down step-by-step, making sure everyone can follow along. So, grab your notebooks, maybe a cup of coffee, and let's get this math party started! Understanding the vertex and x-intercepts gives us a clear picture of where our parabola hits its highest or lowest point (the vertex) and where it crosses the horizontal axis (the x-intercepts). These points are like the landmarks on the graph, telling us crucial information about the function's domain, range, and roots. We'll explore different methods to find these points, ensuring you're equipped with the knowledge to solve similar problems with confidence. Remember, practice makes perfect, and by dissecting this example, you'll be well on your way to mastering quadratic functions. So, let's roll up our sleeves and make sense of $f(x)=9.8 x^2+11.2 x+3.2$ together, rounding to the nearest hundredth where needed. This is going to be fun, guys!

Unpacking the Quadratic Function: $f(x)=9.8 x^2+11.2 x+3.2$

Alright, let's get down to business with our specific quadratic function: $f(x)=9.8 x^2+11.2 x+3.2$. Before we start calculating, it's crucial to recognize the standard form of a quadratic function, which is $f(x) = ax^2 + bx + c$. In our case, $a = 9.8$, $b = 11.2$, and $c = 3.2$. The 'a' value, our $9.8$, tells us a lot right off the bat. Since it's positive, we know our parabola will open upwards, meaning it has a minimum point, which is our vertex. If 'a' were negative, the parabola would open downwards, and the vertex would be a maximum point. The 'b' value, $11.2$, and the 'c' value, $3.2$, also play significant roles in shaping the parabola and determining the locations of its vertex and x-intercepts. The 'c' value, in particular, is easy to spot: it's always the y-intercept, the point where the graph crosses the y-axis. So, for our function, we already know that the graph hits the y-axis at $(0, 3.2)$. Pretty neat, right? Now, let's focus on finding the vertex. The vertex is the most distinctive point on a parabola; it's either the lowest point (if $a>0$) or the highest point (if $a<0$). For our function, since $a=9.8$ is positive, the vertex will be the minimum point. Finding the vertex involves a couple of formulas that are super helpful. The x-coordinate of the vertex is given by the formula $x = -b / (2a)$. Once we have that x-value, we can plug it back into our original function $f(x)$ to find the corresponding y-coordinate. This y-coordinate will be the minimum value of our function. So, for $f(x)=9.8 x^2+11.2 x+3.2$, we'll substitute $a=9.8$ and $b=11.2$ into the formula $x = -b / (2a)$ to find the x-coordinate of our vertex. This initial step is key because it sets us up to find the lowest point of our parabolic curve. Remember, these formulas are your best friends when dealing with quadratics, so make sure you've got them handy. We're going to round our answers to the nearest hundredth, so keep those calculators ready!

Calculating the Vertex: The Lowest Point of Our Parabola

Let's get our hands dirty and calculate the vertex for $f(x)=9.8 x^2+11.2 x+3.2$. As we established, the vertex is the minimum point of our upward-opening parabola (because $a=9.8$ is positive). The formula for the x-coordinate of the vertex is $x = -b / (2a)$. Plugging in our values, where $a = 9.8$ and $b = 11.2$, we get: $x = -(11.2) / (2 * 9.8)$. First, let's calculate the denominator: $2 * 9.8 = 19.6$. So now we have: $x = -11.2 / 19.6$. Now, let's perform the division: $-11.2 / 19.6 ext{ is approximately } -0.57142857...$. The problem asks us to round to the nearest hundredth, so the x-coordinate of our vertex is -0.57. Great job, guys! Now that we have the x-coordinate, we need to find the corresponding y-coordinate. We do this by plugging this x-value back into our original function $f(x)=9.8 x^2+11.2 x+3.2$. So, we'll calculate $f(-0.57)$: $f(-0.57) = 9.8 * (-0.57)^2 + 11.2 * (-0.57) + 3.2$. Let's break this down: $(-0.57)^2 = 0.3249$. Now, multiply that by 9.8: $9.8 * 0.3249 = 3.18402$. Next, calculate $11.2 * (-0.57)$: $11.2 * (-0.57) = -6.384$. Finally, let's put it all together: $f(-0.57) = 3.18402 - 6.384 + 3.2$. Adding and subtracting these values gives us: $3.18402 - 6.384 + 3.2 = 0.00002$. Rounding this to the nearest hundredth, the y-coordinate of our vertex is 0.00. So, the vertex of our quadratic function $f(x)=9.8 x^2+11.2 x+3.2$ is approximately (-0.57, 0.00). This tells us that the lowest point on our parabola is located at x = -0.57, and the minimum value of the function is 0.00. It's pretty cool how these simple formulas can give us such precise information about the graph's shape and position. Keep these calculations handy as we move on to finding the x-intercepts!

Finding the X-Intercepts: Where the Parabola Meets the X-Axis

Now, let's talk about the x-intercepts. These are the points where our function's graph crosses the x-axis. At these points, the y-value (or $f(x)$) is always zero. So, to find the x-intercepts, we need to set our function equal to zero and solve for x: $9.8 x^2+11.2 x+3.2 = 0$. This is a classic quadratic equation, and we have a few ways to solve it. We could try factoring, but with these decimal coefficients, that's likely to be a headache. A more reliable method here is using the quadratic formula, which is: $x = [-b ext{ ± } \sqrt{b^2-4ac}] / (2a)$. This formula is a lifesaver for any quadratic equation, especially when factoring is tough. Let's plug in our values again: $a = 9.8$, $b = 11.2$, and $c = 3.2$. So, the formula becomes: $x = [-11.2 ext{ ± } ext{}\sqrt{(11.2)^2 - 4 * (9.8) * (3.2)}] / (2 * 9.8)$. Let's tackle the parts of this formula step-by-step. First, calculate $b^2$: $(11.2)^2 = 125.44$. Next, calculate $4ac$: $4 * 9.8 * 3.2 = 39.2 * 3.2 = 125.44$. Now, let's find the discriminant, which is the part under the square root: $b^2 - 4ac = 125.44 - 125.44 = 0$. Wow, look at that! The discriminant is 0. This is a special case, guys. When the discriminant ($b^2-4ac$) is exactly zero, it means there is only one unique real solution for x. This also tells us something very important about our parabola: it touches the x-axis at exactly one point. This point is, you guessed it, our vertex! Let's continue with the formula to see this in action: $x = [-11.2 ext{ ± } ext{}\sqrt{0}] / (2 * 9.8)$. Since the square root of 0 is 0, our equation simplifies to: $x = -11.2 / (19.6)$. We already calculated this when finding the x-coordinate of the vertex! $-11.2 / 19.6 ext{ is approximately } -0.57142857...$. Rounding to the nearest hundredth, we get $x ext{ ≈ } -0.57$. So, our quadratic function has only one x-intercept, and it is located at -0.57. This means the parabola is tangent to the x-axis at this point, which aligns perfectly with our vertex calculation where the y-coordinate was 0.00. It's always reassuring when different methods confirm each other, right? This single x-intercept at -0.57 is also the x-coordinate of our vertex, confirming that the minimum point of the parabola lies exactly on the x-axis.

Putting It All Together: The Big Picture

So, let's recap what we've discovered about the quadratic function $f(x)=9.8 x^2+11.2 x+3.2$. We identified that this is a quadratic function in the standard form $f(x) = ax^2 + bx + c$, with $a=9.8$, $b=11.2$, and $c=3.2$. Since $a$ is positive, we knew from the start that the parabola would open upwards, meaning its vertex would be a minimum point. We first tackled the vertex. Using the formula $x = -b / (2a)$, we found the x-coordinate of the vertex to be approximately -0.57. Then, by substituting this x-value back into the function, we calculated the y-coordinate of the vertex to be approximately 0.00. Therefore, the vertex of this parabola is at (-0.57, 0.00). This point signifies the absolute minimum value of the function. Next, we moved on to the x-intercepts, which are the points where the graph crosses the x-axis (where $f(x)=0$). We set up the equation $9.8 x^2+11.2 x+3.2 = 0$ and decided to use the quadratic formula $x = [-b ext{ ± } ext{}\sqrt{b^2-4ac}] / (2a)$ because factoring would be tricky. The magic happened when we calculated the discriminant ($b^2-4ac$), which turned out to be exactly 0. This special result told us that there is only one real solution for x, meaning the parabola touches the x-axis at just one point. This single x-intercept is located at -0.57. This finding perfectly corroborates our vertex calculation; the vertex lies on the x-axis. It's like finding out the lowest point of your roller coaster is exactly at ground level! This is a crucial concept in understanding quadratic functions. The number of x-intercepts (zero, one, or two) is determined by the discriminant: if $b^2-4ac > 0$, there are two distinct x-intercepts; if $b^2-4ac = 0$, there is exactly one x-intercept (and the vertex is on the x-axis); and if $b^2-4ac < 0$, there are no real x-intercepts (the parabola is entirely above or below the x-axis). So, for $f(x)=9.8 x^2+11.2 x+3.2$, we have a parabola that opens upwards, with its lowest point (vertex) touching the x-axis at (-0.57, 0.00). This single point serves as both the vertex and the only x-intercept. Understanding these key features helps us sketch the graph accurately and interpret the behavior of the function in various real-world applications. Keep practicing these techniques, and you'll be a quadratic guru in no time! Stay curious, and happy calculating!