Solving Quadratic Functions And Parabolas: A Step-by-Step Guide

Hey guys! Let's dive into some cool math problems involving quadratic functions and parabolas. We're going to break down how to solve these problems step-by-step, making it super easy to understand. So, grab your pencils and let's get started!

1. Finding the Intersection of a Quadratic Function with the y-axis

Quadratic functions, also known as second-degree polynomials, are all about curves – specifically, parabolas. A parabola is that classic U-shaped curve you see when you graph a quadratic function. Understanding how these functions work is key to solving a bunch of different problems in math and real-world applications. The question we're tackling here asks us to find where a specific quadratic function intersects the y-axis. The y-axis, remember, is the vertical line on a graph where the x-coordinate is always zero.

Understanding the Problem

We're given three points that this quadratic function passes through: (1, -50), (6, 0), and (-4, 0). The crucial thing to understand here is that the equation of a quadratic function can be written in a few different forms. But since we have the x-intercepts (the points where the function crosses the x-axis) we can use a special form, which is super helpful here. This form is called the factored form. Because we know that our function passes through (6, 0) and (-4, 0), we know these are the x-intercepts. So, the factored form of the quadratic equation will look like this: y = a(x - 6)(x + 4), where a is a constant we need to find.

Solving for the Constant a

To find the value of a, we use the third point we're given, (1, -50). This point tells us that when x = 1, y = -50. Substitute these values into the equation: -50 = a(1 - 6)(1 + 4), which simplifies to -50 = a(-5)(5), and even further to -50 = -25a. Then, to isolate a, divide both sides by -25: a = 2. So now we know the full equation is y = 2(x - 6)(x + 4).

Finding the y-intercept

To find the y-intercept, we need to know what the value of y is when x = 0. So, we substitute x = 0 into our equation: y = 2(0 - 6)(0 + 4), which simplifies to y = 2(-6)(4), and finally y = -48. That means the quadratic function intersects the y-axis at the point (0, -48). Therefore, the correct answer is (B) (0, -48). See? Not so bad, right?

This is a classic example of how understanding the different forms of quadratic equations—especially the factored form—can make solving problems a whole lot easier. Plus, knowing how to substitute values and solve for unknowns are skills that are useful across all sorts of math problems, and indeed, in life.

2. Determining the Extremum of a Parabola

Alright, let's switch gears and talk about parabolas. The extremum of a parabola refers to its vertex – the point where the parabola reaches its highest or lowest value. Because of the shape of the parabola, this extremum point is either a minimum (if the parabola opens upwards) or a maximum (if the parabola opens downwards). The given equation is y = x^2 + x + 2.

Understanding the Equation

This equation is in the standard form of a quadratic equation: y = ax² + bx + c. In our equation, a = 1, b = 1, and c = 2. Because the coefficient a is positive (it's 1 in this case), the parabola opens upwards. This means the vertex is the minimum point on the curve.

Finding the x-coordinate of the Vertex

The x-coordinate of the vertex can be found using the formula: x = -b / (2a). Substituting our values, we get x = -1 / (2 * 1) = -1/2. This is the x-coordinate of our vertex.

Finding the y-coordinate of the Vertex (The Extreme Value)

To find the y-coordinate (which is the extreme value we are looking for), substitute the x-coordinate back into the original equation. So, y = (-1/2)² + (-1/2) + 2. This simplifies to y = 1/4 - 1/2 + 2, which then becomes y = 1/4 - 2/4 + 8/4, and finally, y = 7/4. Therefore, the vertex of the parabola is at (-1/2, 7/4). The extreme value (the y-coordinate of the vertex) is 7/4. Since the parabola opens upward, this is a minimum value.

Conclusion

So, the minimum value of the parabola y = x² + x + 2 is 7/4. It is essential to recognize the link between the coefficient 'a' and the direction the parabola opens. If 'a' is positive, the parabola opens upwards, and if 'a' is negative, it opens downwards. This tells you instantly whether you are dealing with a minimum or a maximum value. Keep practicing, and these concepts will become second nature, I promise!

This method is a core skill when dealing with parabolas and understanding the vertex form. Recognizing the relationship between the coefficients of the equation and the shape/orientation of the parabola is also key to efficiently solving these types of problems.

I hope this step-by-step guide was helpful, guys! Keep practicing, and you'll become a pro at these quadratic function and parabola problems in no time! Remember, math is all about practice and understanding the underlying concepts. Good luck, and have fun!