Finding The Equation Of A Parabola With Vertex (6,-3) And Point (2,-2)

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Hey guys! Today, we're diving into the fascinating world of parabolas and tackling a classic problem: finding the equation of a parabola given its vertex and a point it passes through. This is a fundamental concept in algebra and precalculus, and mastering it will unlock a deeper understanding of quadratic functions and their graphical representations. So, let's put on our mathematical hats and get started!

Understanding Parabolas: A Quick Recap

Before we jump into the problem, let's refresh our understanding of parabolas. A parabola is a symmetrical, U-shaped curve. It's defined as the set of all points that are equidistant to a fixed point (the focus) and a fixed line (the directrix). Parabolas pop up everywhere in the real world, from the trajectory of a ball thrown in the air to the shape of satellite dishes and the reflectors in car headlights.

The most general form of a parabola's equation is the quadratic equation: y = ax^2 + bx + c. However, when dealing with the vertex of a parabola, it's often more convenient to use the vertex form of the equation. For a parabola that opens vertically (upwards or downwards), the vertex form is:

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

Where:

  • (h, k) represents the coordinates of the vertex.
  • 'a' determines the direction and the width of the parabola. If 'a' is positive, the parabola opens upwards; if 'a' is negative, it opens downwards. The larger the absolute value of 'a', the narrower the parabola.

But what about parabolas that open horizontally (leftwards or rightwards)? Well, in that case, we simply swap the roles of x and y in the equation. So, the vertex form for a horizontally opening parabola is:

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

Notice that the roles of h and k remain the same – (h, k) still represents the vertex. The sign of 'a' still determines the direction of opening: positive for rightwards, negative for leftwards. And, as before, the magnitude of 'a' affects the width of the parabola.

In this problem, the given options for the equation of the parabola are in the form x = ay^2 + by + c. This indicates that we're dealing with a parabola that opens either to the left or to the right. This is a crucial observation that guides our problem-solving approach.

Problem Breakdown: Vertex (6,-3) and Point (2,-2)

Now that we've refreshed our knowledge of parabolas, let's break down the problem at hand. We're given two key pieces of information:

  1. The vertex of the parabola is at the point (6, -3).
  2. The parabola passes through the point (2, -2).

Our mission is to find the equation of the parabola that satisfies these conditions. We're given four options, and only one of them is the correct answer. So, we need a systematic way to determine which equation fits the bill.

The first piece of information, the vertex (6, -3), is incredibly valuable. It allows us to immediately write down the vertex form of the parabola's equation, with the only unknown being the value of 'a'. Since the options are in the form x = ay^2 + by + c, we know we should use the vertex form for a horizontally opening parabola:

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

Substituting the vertex coordinates (h, k) = (6, -3), we get:

x = a(y - (-3))^2 + 6

Simplifying, this becomes:

x = a(y + 3)^2 + 6

This is a significant step forward! We've narrowed down the possibilities to a single unknown, 'a'. This constant determines the parabola's direction and width, and it's the key to unlocking the complete equation.

Solving for 'a': Using the Point (2,-2)

To find the value of 'a', we'll use the second piece of information: the parabola passes through the point (2, -2). This means that the coordinates (x, y) = (2, -2) must satisfy the equation of the parabola. In other words, if we plug in these values for x and y in the equation we derived above, the equation must hold true.

So, let's substitute x = 2 and y = -2 into the equation:

2 = a(-2 + 3)^2 + 6

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

2 = a(1)^2 + 6 2 = a + 6 a = 2 - 6 a = -4

Ta-da! We've found the value of 'a'. It's -4. This tells us that the parabola opens to the left (since 'a' is negative) and that it's relatively narrow (since the absolute value of 'a' is 4).

The Equation Unveiled: Putting It All Together

Now that we know the value of 'a', we can substitute it back into the vertex form of the equation:

x = a(y + 3)^2 + 6

x = -4(y + 3)^2 + 6

This is the equation of the parabola in vertex form. However, the options given in the problem are in the general form x = ay^2 + by + c. So, we need to expand the equation and rewrite it in that form.

Let's expand the squared term:

x = -4(y^2 + 6y + 9) + 6

Now, distribute the -4:

x = -4y^2 - 24y - 36 + 6

Finally, combine the constant terms:

x = -4y^2 - 24y - 30

And there we have it! This is the equation of the parabola in the general form. It matches one of the options provided in the problem. Let's take a look at the options again to confirm.

Solution and Verification

We've arrived at the equation:

x = -4y^2 - 24y - 30

Now, let's compare this with the options given in the problem:

A. x = -4y^2 - 24y - 42 B. x = 4y^2 - 24y + 42 C. x = -16y^2 - 48y - 138 D. x = -4y^2 - 24y - 30

We can clearly see that our equation matches option D. So, the correct answer is:

D. x = -4y^2 - 24y - 30

We've successfully found the equation of the parabola! But let's not stop here. It's always a good idea to verify our answer to make sure we haven't made any mistakes along the way.

We can verify our solution in a couple of ways:

  1. Check the vertex: We can rewrite our equation in vertex form to ensure that the vertex is indeed (6, -3). This is essentially the reverse of the process we did earlier. By completing the square or using the vertex formula, we can confirm that the vertex form is x = -4(y + 3)^2 + 6, which confirms the vertex.

  2. Check the point (2, -2): We can plug in x = 2 and y = -2 into our equation to see if it holds true:

    2 = -4(-2)^2 - 24(-2) - 30 2 = -16 + 48 - 30 2 = 2

    The equation holds true! This gives us further confidence that our solution is correct.

Key Takeaways: Mastering Parabola Equations

We've successfully navigated through this parabola problem, but more importantly, we've reinforced some key concepts and problem-solving techniques. Let's recap the essential takeaways:

  • Vertex form is your friend: When dealing with problems involving the vertex of a parabola, the vertex form of the equation (y = a(x - h)^2 + k or x = a(y - k)^2 + h) is often the most convenient starting point. It directly incorporates the vertex coordinates, making the problem much simpler.
  • Use given information strategically: Each piece of information provided in the problem is crucial. In this case, the vertex gave us the (h, k) values for the vertex form, and the point (2, -2) allowed us to solve for the unknown constant 'a'.
  • Don't forget the direction of opening: Pay attention to whether the parabola opens vertically or horizontally. This determines which form of the vertex equation you should use.
  • Verify your solution: Always take the time to verify your answer, especially in exams or assessments. This can catch any minor errors and ensure that you get the correct result.

Practice Makes Perfect: Sharpen Your Parabola Skills

Congratulations, guys! You've conquered this parabola problem and gained valuable insights into working with quadratic equations and their graphs. But remember, mathematics is a skill that improves with practice. To truly master parabola equations, try solving similar problems with different vertex points and other given points. You can also explore problems that involve finding the focus and directrix of a parabola. The more you practice, the more confident and proficient you'll become!

Keep exploring, keep learning, and keep those mathematical gears turning!