Finding Roots And Vertex Of G(t) = T² - T - 42 A Comprehensive Guide

Hey guys! Let's dive into the fascinating world of quadratic functions. Today, we're going to dissect the function g(t) = t² - t - 42. We'll find its roots (where the function crosses the x-axis) and pinpoint the vertex, which is the highest or lowest point on the parabola. Understanding these elements gives us a complete picture of the function's behavior.

1) Unveiling the Roots of g(t) = t² - t - 42

Finding the roots of a quadratic function is like discovering the treasure hidden within its equation. The roots, also known as the zeros or x-intercepts, are the values of t that make the function g(t) equal to zero. In simpler terms, these are the points where the parabola intersects the horizontal axis on our graph. To find these crucial points, we need to solve the equation t² - t - 42 = 0. There are several methods to accomplish this mission, but the most common ones are factoring, using the quadratic formula, or completing the square.

Let's start with factoring, as it's often the quickest route if the quadratic expression cooperates. We are looking for two numbers that multiply to -42 (the constant term) and add up to -1 (the coefficient of the t term). After some mental gymnastics, we find that -7 and 6 fit the bill perfectly. So, we can rewrite our equation as (t - 7)(t + 6) = 0. Now, the magic happens: for the product of two factors to be zero, at least one of them must be zero. This gives us two possibilities: t - 7 = 0 or t + 6 = 0. Solving these simple equations, we get t = 7 and t = -6. So, the roots of our function are -6 and 7. Remember, we need to write the smaller t value first, followed by the larger one. Therefore, the roots are smaller t = -6 and larger t = 7.

But what if factoring isn't straightforward? Fear not! The quadratic formula is our trusty backup plan. This formula works for any quadratic equation in the form at² + bt + c = 0. The formula is t = (-b ± √(b² - 4ac)) / 2a. In our case, a = 1, b = -1, and c = -42. Plugging these values into the formula, we get t = (1 ± √((-1)² - 4 * 1 * -42)) / (2 * 1). Simplifying, we have t = (1 ± √(1 + 168)) / 2, which becomes t = (1 ± √169) / 2. Since √169 is 13, we get t = (1 ± 13) / 2. This gives us two solutions: t = (1 + 13) / 2 = 7 and t = (1 - 13) / 2 = -6. Notice, we arrived at the same roots as before, confirming our factoring method was correct! The roots -6 and 7 are crucial as they define where the parabola intersects the t-axis, providing key anchor points for sketching the graph and understanding the function’s behavior.

Understanding the roots is fundamental in many real-world applications. For example, in physics, if g(t) represents the height of a projectile at time t, the roots tell us when the projectile hits the ground. In business, if g(t) represents profit, the roots tell us the break-even points. So, grasping how to find and interpret roots is super useful. We’ve nailed this part, guys! Now let's move on to finding the vertex, which will give us even more insight into our function.

2) Pinpointing the Vertex of the Parabola for g(t) = t² - t - 42

The vertex of a parabola is its most extreme point – either the lowest point (minimum) if the parabola opens upwards or the highest point (maximum) if it opens downwards. For our function, g(t) = t² - t - 42, the coefficient of the t² term is positive (it's 1), which means the parabola opens upwards, and the vertex will be the minimum point. Finding the vertex is like finding the sweet spot of our function, the point where it changes direction.

There are a couple of ways to find the vertex. One method involves using a formula, and the other involves completing the square. Let's start with the formula, as it's often the quickest way to get to the vertex coordinates. The t-coordinate of the vertex (t_vertex) is given by the formula t_vertex = -b / 2a. Remember from our quadratic formula discussion, a is the coefficient of t² and b is the coefficient of t. In our case, a = 1 and b = -1. Plugging these values into the formula, we get t_vertex = -(-1) / (2 * 1) = 1 / 2 = 0.5. So, the t-coordinate of the vertex is 0.5. This tells us the horizontal position of the vertex on our graph.

Now that we have the t-coordinate, we need to find the g(t)-coordinate, which is the vertical position of the vertex. To do this, we simply plug t_vertex back into our original function: g(0.5) = (0.5)² - 0.5 - 42. This simplifies to g(0.5) = 0.25 - 0.5 - 42 = -42.25. Therefore, the vertex of the parabola is (0.5, -42.25). This means the lowest point on the parabola is at t = 0.5, and the function value at that point is -42.25.

Alternatively, we can find the vertex by completing the square. This method involves rewriting the quadratic expression in vertex form, which is g(t) = a(t - h)² + k, where (h, k) is the vertex. To complete the square for g(t) = t² - t - 42, we first focus on the t² - t part. We need to add and subtract a value that will make this a perfect square trinomial. The value we need to add and subtract is (b / 2a)², which in our case is (-1 / (2 * 1))² = (0.5)² = 0.25. So, we rewrite the function as g(t) = t² - t + 0.25 - 0.25 - 42. The first three terms now form a perfect square: (t - 0.5)². So, our function becomes g(t) = (t - 0.5)² - 0.25 - 42, which simplifies to g(t) = (t - 0.5)² - 42.25. Comparing this to the vertex form, we see that h = 0.5 and k = -42.25, confirming that the vertex is indeed (0.5, -42.25).

Knowing the vertex is super helpful because it tells us the minimum (or maximum) value of the function and where it occurs. In practical scenarios, this could represent the minimum cost, the maximum profit, or the lowest point in a physical trajectory. Plus, the vertex helps us sketch the parabola accurately. We know it opens upwards, and we now know its lowest point. So, finding the vertex is another key piece in understanding our quadratic function. We’ve explored the roots and the vertex – that’s a solid understanding of g(t)!

Wrapping Up Our Quadratic Journey

Alright, guys, we’ve taken a deep dive into the quadratic function g(t) = t² - t - 42. We successfully found the roots by factoring and using the quadratic formula, and we pinpointed the vertex using both the formula and completing the square method. We discovered that the roots are t = -6 and t = 7, and the vertex is located at (0.5, -42.25). This information gives us a comprehensive understanding of the function’s behavior and its graphical representation.

Remember, quadratic functions are everywhere, from the trajectory of a ball to the design of suspension bridges. Mastering these concepts not only helps us in math class but also equips us with tools to understand and solve real-world problems. So keep practicing, keep exploring, and keep having fun with math! You’ve got this!

Now you have a solid grasp of how to analyze a quadratic function. Go forth and conquer more mathematical challenges!