The Vertex Of This Parabola Is At 2

Let's delve into the fascinating world of parabolas, exploring what it means when we say "the vertex of this parabola is at 2." This statement, while seemingly simple, unlocks a wealth of information about the parabola's behavior, equation, and graphical representation. We'll unravel the meaning of the vertex, its significance, and how to work with parabolas when the vertex has a specific x-coordinate.

Understanding the Parabola and Its Vertex

A parabola is a symmetrical, U-shaped curve. It's formally defined as the set of all points equidistant to a fixed point (the focus) and a fixed line (the directrix). Parabolas appear in various contexts, from the path of a projectile to the design of satellite dishes.

The vertex is the turning point of the parabola. It's the point where the parabola changes direction. This point is crucial because:

• It represents either the minimum or maximum value of the quadratic function that defines the parabola.
• It's the point of symmetry for the entire curve. A vertical line drawn through the vertex (the axis of symmetry) divides the parabola into two mirror images.
• Knowing the vertex helps us easily graph the parabola.

When we say "the vertex of this parabola is at 2," we're specifically referring to the x-coordinate of the vertex. This tells us the horizontal position of the turning point. The complete vertex is a coordinate pair (h, k), where 'h' is the x-coordinate and 'k' is the y-coordinate. So, in this case, h = 2. We still need to determine the y-coordinate (k) to fully define the vertex.

Forms of a Parabola's Equation

To understand how the x-coordinate of the vertex influences the parabola's equation, let's review the different forms of a quadratic equation:

1. Standard Form: y = ax² + bx + c
   • 'a', 'b', and 'c' are constants, and 'a' cannot be zero.
   • While the standard form is useful for various calculations, the vertex is not immediately apparent.
2. Vertex Form: y = a(x - h)² + k
   • 'a' is the same coefficient as in the standard form.
   • '(h, k)' represents the coordinates of the vertex. This is the most useful form when you know the vertex.
3. Factored Form: y = a(x - r₁)(x - r₂)
   • 'a' is again the same coefficient.
   • 'r₁' and 'r₂' are the x-intercepts (roots or zeros) of the parabola.
   • The vertex can be found by averaging the roots: h = (r₁ + r₂) / 2, then substituting this 'h' value back into the equation to find 'k'.

Working with the Vertex at x = 2

Now, let's focus on the scenario where the x-coordinate of the vertex (h) is equal to 2. This means our vertex has the form (2, k), where 'k' is still unknown. Here's how this information impacts the different forms of the equation:

1. Vertex Form:

This is the most straightforward case. Since we know h = 2, we can directly substitute it into the vertex form equation:

• y = a(x - 2)² + k

To fully define the equation, we need to determine the values of 'a' and 'k'. Let's look at ways to find these:

• If we are given another point on the parabola: Substitute the x and y coordinates of that point into the equation y = a(x - 2)² + k. This gives you one equation with two unknowns ('a' and 'k'). You'll need another piece of information (like another point, or a relationship between 'a' and 'k') to solve for both variables.
• If we are given the y-coordinate of the vertex (k): We now know the complete vertex (2, k), and the only unknown is 'a'. If we also have another point on the parabola, we can substitute its coordinates into the equation y = a(x - 2)² + k and solve for 'a'.
• If we are given the focus and directrix: The vertex is the midpoint between the focus and the directrix. Since we know the x-coordinate of the vertex is 2, and we know the relationship between the focus, directrix, and vertex, we can potentially solve for the y-coordinate of the vertex (k) and the value of 'a'. (Recall that the distance from the vertex to the focus is equal to the distance from the vertex to the directrix).

Example:

Suppose we know the vertex is at x = 2, and the parabola also passes through the point (3, 5). Let's also say we know that the y-coordinate of the vertex is k = 1. Then our vertex is (2, 1). We can now find 'a':

• y = a(x - 2)² + 1
• Substitute the point (3, 5): 5 = a(3 - 2)² + 1
• Simplify: 5 = a(1)² + 1
• 5 = a + 1
• a = 4

Therefore, the equation of the parabola is y = 4(x - 2)² + 1.

2. Standard Form:

If we want to express the parabola in standard form (y = ax² + bx + c), knowing that the vertex is at x = 2 can still be helpful. Here's how we can approach it:

• Convert from Vertex Form: If you've already found the equation in vertex form (as shown above), you can simply expand the expression to get it into standard form. For example, using the equation we found earlier:

  • y = 4(x - 2)² + 1
  • y = 4(x² - 4x + 4) + 1
  • y = 4x² - 16x + 16 + 1
  • y = 4x² - 16x + 17

  So, in this case, a = 4, b = -16, and c = 17.

• Using the formula h = -b / 2a: The x-coordinate of the vertex (h) is related to the coefficients 'a' and 'b' in the standard form equation by the formula h = -b / 2a. Since we know h = 2, we have:

  • 2 = -b / 2a
  • 4a = -b
  • b = -4a

  This tells us that 'b' must be -4 times the value of 'a'. However, this alone isn't enough to determine the equation. We would need another piece of information, such as a point on the parabola, or the y-intercept (the value of 'c').

  Example:

  Let's say we also know that the y-intercept is 17 (i.e., the parabola passes through the point (0, 17)). We now have:

  • y = ax² + bx + c
  • y = ax² - 4ax + c (Substituting b = -4a)
  • 17 = a(0)² - 4a(0) + c (Substituting the point (0, 17))
  • c = 17

  Now we have y = ax² - 4ax + 17. We still need another point on the parabola to find the value of 'a', or we could use the fact that the vertex is at (2, k) and find 'k' by substituting x = 2:

  • k = a(2)² - 4a(2) + 17
  • k = 4a - 8a + 17
  • k = -4a + 17

  If we knew the value of 'k', we could easily solve for 'a'. Alternatively, if we had another point on the parabola (besides (0, 17)), we could substitute that in to y = ax² - 4ax + 17 and solve for 'a'. As you can see, working backwards from the standard form can be more involved.

3. Factored Form:

Knowing that the vertex is at x = 2 can be useful in conjunction with the factored form (y = a(x - r₁)(x - r₂)) if you also know one of the x-intercepts (roots).

• Relationship between Vertex and Roots: The x-coordinate of the vertex is the average of the two roots: h = (r₁ + r₂) / 2. If we know h = 2, and we know one of the roots (say, r₁), we can find the other root (r₂).

  • 2 = (r₁ + r₂) / 2
  • 4 = r₁ + r₂
  • r₂ = 4 - r₁

  So, if you knew that one root was, say, r₁ = 1, then the other root would be r₂ = 4 - 1 = 3. The factored form of the equation would then be y = a(x - 1)(x - 3). You would still need another point on the parabola to determine the value of 'a'.

Example:

Suppose the parabola has a vertex at x = 2 and x-intercepts at x = 1 and x = 3. To find the equation:

• y = a(x - 1)(x - 3)
• Then, let's say we also know the parabola passes through the point (0,3).
• 3 = a (0-1)(0-3)
• 3 = 3a
• a = 1

Therefore, y = (x-1)(x-3).

Illustrative Examples and Applications

Let's solidify our understanding with some more examples:

Example 1: Finding the Equation Given Vertex and a Point

A parabola has its vertex at (2, -3) and passes through the point (0, 1). Find the equation of the parabola in vertex form and standard form.

• Vertex Form: We know h = 2 and k = -3, so the vertex form is y = a(x - 2)² - 3.
• To find 'a', substitute the point (0, 1): 1 = a(0 - 2)² - 3
• 1 = 4a - 3
• 4 = 4a
• a = 1
• Therefore, the equation in vertex form is y = (x - 2)² - 3.
• Standard Form: Expand the vertex form: y = (x² - 4x + 4) - 3
• y = x² - 4x + 1

Example 2: Maximizing Profit

A company's profit (P) is modeled by the quadratic equation P = -2x² + 8x + 5, where 'x' is the number of units sold. Find the number of units that maximizes profit.

• The profit function is a parabola opening downwards (because a = -2, which is negative). The maximum profit occurs at the vertex.
• The x-coordinate of the vertex is given by x = -b / 2a. In this case, a = -2 and b = 8.
• x = -8 / (2 * -2) = -8 / -4 = 2
• Therefore, the company maximizes its profit when it sells 2 units. Notice that we didn't even need to find the y-coordinate of the vertex (which would represent the maximum profit), because the question only asked for the number of units that maximizes profit.

Example 3: Projectile Motion

The height (h) of a projectile is given by the equation h = -16t² + 64t + 80, where 't' is the time in seconds. Find the time at which the projectile reaches its maximum height.

• This is another parabola opening downwards (a = -16). The maximum height occurs at the vertex.
• t = -b / 2a = -64 / (2 * -16) = -64 / -32 = 2
• The projectile reaches its maximum height at t = 2 seconds.

Common Mistakes to Avoid

• Confusing 'h' and 'k': Remember that 'h' is the x-coordinate of the vertex, and 'k' is the y-coordinate.
• Incorrectly applying the formula h = -b / 2a: Double-check your signs when calculating the x-coordinate of the vertex.
• Forgetting the 'a' value: The coefficient 'a' determines the parabola's width and direction (opening upwards or downwards). Don't neglect it!
• Assuming the vertex is always at the origin: The vertex can be anywhere on the coordinate plane. The statement "the vertex of this parabola is at 2" specifically tells you the x-coordinate of the vertex.
• Not using enough information: To fully define a parabola, you typically need three pieces of independent information (e.g., the vertex and another point, two points and the 'a' value, etc.).

Conclusion

Knowing that the x-coordinate of a parabola's vertex is a specific value (like 2) provides a significant starting point for understanding and defining the parabola. By understanding the different forms of a quadratic equation (vertex form, standard form, factored form) and how the vertex relates to each form, we can effectively solve problems, graph parabolas, and apply this knowledge to real-world scenarios involving optimization and projectile motion. Remember to carefully analyze the given information and choose the most appropriate form of the equation to make the problem-solving process more efficient. With practice and a solid understanding of the fundamentals, you'll be able to confidently tackle any parabola-related challenge.