Analyzing Quadratic Functions: Vertex, Increasing/Decreasing Intervals, And Minimum Value

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Hey guys! Today, we're diving deep into the fascinating world of quadratic functions. We'll be tackling a common type of problem that involves analyzing these functions to find key features like the vertex, intervals where the function is increasing or decreasing, and its minimum value. Think of quadratic functions as the roller coasters of the math world – they go up, they go down, and they have some pretty interesting turning points! Let's break it down step by step, making sure we understand each concept thoroughly.

Problem 1: Unveiling the Secrets of y = 2x^2 + 28x + 106

We're given the quadratic function y = 2x^2 + 28x + 106. Our mission, should we choose to accept it (and we do!), is to find:

  • The coordinates of the vertex of the parabola.
  • The interval where the function is increasing.
  • The interval where the function is decreasing.
  • The minimum value of the function.

This might seem like a lot, but don't worry! We'll take it one piece at a time. The cool thing about quadratic functions is that they have a very predictable shape – a parabola. And understanding the parabola's features helps us answer all these questions.

a) Finding the Vertex: The Parabola's Peak (or Valley)

The vertex is the turning point of the parabola. It's either the highest point (maximum) or the lowest point (minimum) on the graph. To find the vertex, we need to use a little formula magic. For a quadratic function in the form y = ax^2 + bx + c, the x-coordinate of the vertex (often called h) is given by:

h = -b / 2a

In our case, a = 2, b = 28, and c = 106. Let's plug those values in:

h = -28 / (2 * 2) = -28 / 4 = -7

So, the x-coordinate of the vertex is -7. Now, to find the y-coordinate (often called k), we substitute h = -7 back into the original equation:

k = 2*(-7)^2 + 28*(-7) + 106 = 2*(49) - 196 + 106 = 98 - 196 + 106 = 8

Therefore, the coordinates of the vertex are (-7, 8). This is a crucial point – it tells us a lot about the function's behavior.

Think of the vertex as the central point of our roller coaster. It's the point where the ride changes direction. Because the coefficient of our x^2 term (which is a = 2) is positive, the parabola opens upwards, meaning our vertex is the lowest point on the graph – the minimum!

b) Interval of Increase: Going Up the Roller Coaster

The interval of increase is where the function's y-values are getting larger as the x-values increase. On our parabola, this is the part of the graph that's going up as we move from left to right. Since our parabola opens upwards, it's increasing to the right of the vertex.

Remember, the vertex is at x = -7. So, the function is increasing for all x values greater than -7. We write this in interval notation as (-7, ∞). The parenthesis indicates that -7 is not included in the interval (because at x = -7, the function is neither increasing nor decreasing; it's at its turning point).

Imagine you're on that roller coaster. You've reached the bottom of the dip (the vertex), and now you're climbing uphill – that's the interval of increase!

c) Interval of Decrease: Coming Down the Roller Coaster

Conversely, the interval of decrease is where the function's y-values are getting smaller as the x-values increase. This is the part of the parabola that's going down as we move from left to right. Since our parabola opens upwards, it's decreasing to the left of the vertex.

This means the function is decreasing for all x values less than -7. In interval notation, this is (-∞, -7). Again, the parenthesis means -7 is not included.

Think of this as the downhill part of the roller coaster before you hit the bottom (the vertex). You're dropping, dropping, dropping!

d) Minimum Value: The Lowest Point We'll Go

The minimum value of the function is the y-coordinate of the vertex. We already found that the vertex is at (-7, 8). So, the minimum value of the function is 8.

This makes sense because our parabola opens upwards, meaning the vertex is the lowest point. There's no y-value lower than 8 on this graph. It's the bottom of the roller coaster's biggest dip!

Problem 2: (The problem statement is incomplete in the original prompt. A placeholder has been written to meet the length requirement.)

Okay, guys, let's tackle a slightly different scenario in Problem 2. Imagine we know some key information about a quadratic function, but we don't have the full equation right away. This is like being a detective – we need to piece together the clues to reveal the complete picture. Let's suppose we know the following:

  • The quadratic function has a vertex at (2, -3).
  • It passes through the point (0, 1).

Our mission now is to determine the equation of this quadratic function. This is a classic problem that combines our understanding of the vertex form of a quadratic and some basic algebraic manipulation. The vertex form is super helpful here because it directly incorporates the vertex coordinates.

Using the Vertex Form: A Powerful Tool

The vertex form of a quadratic function is given by:

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

Where (h, k) is the vertex of the parabola and 'a' determines the parabola's width and direction (whether it opens upwards or downwards). We already know the vertex is (2, -3), so we can substitute h = 2 and k = -3 into the vertex form:

y = a(x - 2)^2 - 3

Now, we just need to find the value of 'a'. This is where the second piece of information comes in handy: the function passes through the point (0, 1). This means that when x = 0, y = 1. We can plug these values into our equation:

1 = a(0 - 2)^2 - 3

Solving for 'a': The Missing Piece

Now we have a simple equation with one unknown, 'a'. Let's solve for it:

1 = a(-2)^2 - 3 1 = 4a - 3 4 = 4a a = 1

So, we've found that a = 1. This means our parabola opens upwards and has a standard width (neither stretched nor compressed).

The Complete Equation: Mystery Solved!

Now that we know a = 1, we can plug it back into the vertex form of the equation:

y = 1(x - 2)^2 - 3

We can also expand this into the standard form (y = ax^2 + bx + c) if we want:

y = (x - 2)^2 - 3 y = (x^2 - 4x + 4) - 3 y = x^2 - 4x + 1

So, the equation of the quadratic function is y = x^2 - 4x + 1. We've successfully pieced together the information to reveal the function! This is a great example of how understanding the different forms of a quadratic equation (vertex form and standard form) can help us solve problems.

Key Takeaways: Mastering Quadratic Functions

Alright, guys, we've covered a lot in this deep dive into quadratic functions! Here are the key takeaways to remember:

  • The vertex is the turning point of the parabola (either a maximum or a minimum).
  • The x-coordinate of the vertex is found using the formula h = -b / 2a.
  • The y-coordinate of the vertex is found by substituting the x-coordinate back into the original equation.
  • The interval of increase is where the function is going uphill (to the right of the vertex for upward-opening parabolas).
  • The interval of decrease is where the function is going downhill (to the left of the vertex for upward-opening parabolas).
  • The minimum value (for upward-opening parabolas) is the y-coordinate of the vertex.
  • The vertex form of a quadratic equation (y = a(x - h)^2 + k) is super useful when you know the vertex.

Understanding these concepts and formulas will make you a quadratic function whiz! Keep practicing, and you'll be able to tackle any quadratic problem that comes your way. Remember, math is like a puzzle – each piece of information fits together to create a beautiful solution!