Solve The Equation:$\[ Y = \frac{1}{2}(x + 6)^2 - 8 \\]

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Introduction

In this article, we will delve into the world of mathematics and explore a method for solving a quadratic equation. The equation we will be working with is y=12(x+6)28y = \frac{1}{2}(x + 6)^2 - 8. This equation is a quadratic equation in the form of y=ax2+bx+cy = ax^2 + bx + c, where aa, bb, and cc are constants. Our goal is to isolate the variable xx and find its value.

Understanding the Equation

Before we begin solving the equation, let's take a closer look at its structure. The equation is in the form of y=12(x+6)28y = \frac{1}{2}(x + 6)^2 - 8. We can see that the equation is a quadratic equation, with a squared term (x+6)2(x + 6)^2 and a constant term 8-8. The coefficient of the squared term is 12\frac{1}{2}, which means that the parabola will be wider than a standard parabola.

Expanding the Equation

To solve the equation, we need to expand the squared term. We can do this by using the formula (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2. In this case, a=xa = x and b=6b = 6. Therefore, we can expand the squared term as follows:

12(x+6)2=12(x2+12x+36)\frac{1}{2}(x + 6)^2 = \frac{1}{2}(x^2 + 12x + 36)

Now, we can simplify the equation by distributing the 12\frac{1}{2} to the terms inside the parentheses:

y=12x2+6x+188y = \frac{1}{2}x^2 + 6x + 18 - 8

Combining Like Terms

We can now combine the like terms on the right-hand side of the equation. The terms 6x6x and 8-8 are like terms, so we can combine them as follows:

y=12x2+6x+10y = \frac{1}{2}x^2 + 6x + 10

Solving for x

Now that we have simplified the equation, we can solve for xx. To do this, we need to isolate the variable xx on one side of the equation. We can do this by subtracting 12x2\frac{1}{2}x^2 from both sides of the equation:

12x2+6x+1012x2=y\frac{1}{2}x^2 + 6x + 10 - \frac{1}{2}x^2 = y

This simplifies to:

6x+10=y6x + 10 = y

Next, we can subtract 1010 from both sides of the equation:

6x=y106x = y - 10

Finally, we can divide both sides of the equation by 66 to solve for xx:

x=y106x = \frac{y - 10}{6}

Conclusion

In this article, we have solved the quadratic equation y=12(x+6)28y = \frac{1}{2}(x + 6)^2 - 8. We expanded the squared term, combined like terms, and solved for xx. The final solution is x=y106x = \frac{y - 10}{6}. This equation can be used to find the value of xx for any given value of yy.

Graphing the Equation

To visualize the equation, we can graph it on a coordinate plane. The graph of the equation will be a parabola that opens upwards. The vertex of the parabola will be at the point (x,y)=(6,8)(x, y) = (-6, -8).

Real-World Applications

The equation y=12(x+6)28y = \frac{1}{2}(x + 6)^2 - 8 has many real-world applications. For example, it can be used to model the trajectory of a projectile, such as a thrown ball or a rocket. It can also be used to model the growth of a population over time.

Final Thoughts

Introduction

In our previous article, we solved the quadratic equation y=12(x+6)28y = \frac{1}{2}(x + 6)^2 - 8. We expanded the squared term, combined like terms, and solved for xx. In this article, we will answer some frequently asked questions about solving quadratic equations.

Q: What is a quadratic equation?

A quadratic equation is a polynomial equation of degree two, which means that the highest power of the variable is two. The general form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0, where aa, bb, and cc are constants.

Q: How do I know if an equation is quadratic?

To determine if an equation is quadratic, look for the highest power of the variable. If the highest power is two, then the equation is quadratic. For example, the equation x2+3x+2=0x^2 + 3x + 2 = 0 is quadratic because the highest power of xx is two.

Q: What is the difference between a quadratic equation and a linear equation?

A quadratic equation is a polynomial equation of degree two, while a linear equation is a polynomial equation of degree one. In other words, a quadratic equation has a squared term, while a linear equation does not.

Q: How do I solve a quadratic equation?

To solve a quadratic equation, you can use the following steps:

  1. Expand the squared term.
  2. Combine like terms.
  3. Solve for xx.

Q: What is the formula for solving a quadratic equation?

The formula for solving a quadratic equation is:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

This formula is known as the quadratic formula.

Q: What is the significance of the quadratic formula?

The quadratic formula is a powerful tool for solving quadratic equations. It allows us to find the solutions to quadratic equations in a simple and efficient way.

Q: Can I use the quadratic formula to solve all quadratic equations?

Yes, the quadratic formula can be used to solve all quadratic equations. However, it's worth noting that the quadratic formula may not always give you the simplest solution.

Q: What are some common mistakes to avoid when solving quadratic equations?

Some common mistakes to avoid when solving quadratic equations include:

  • Not expanding the squared term.
  • Not combining like terms.
  • Not solving for xx correctly.

Q: How do I graph a quadratic equation?

To graph a quadratic equation, you can use the following steps:

  1. Find the vertex of the parabola.
  2. Find the x-intercepts of the parabola.
  3. Plot the points on a coordinate plane.

Q: What are some real-world applications of quadratic equations?

Quadratic equations have many real-world applications, including:

  • Modeling the trajectory of a projectile.
  • Modeling the growth of a population over time.
  • Finding the maximum or minimum value of a function.

Conclusion

In this article, we have answered some frequently asked questions about solving quadratic equations. We have discussed the definition of a quadratic equation, the difference between a quadratic equation and a linear equation, and the formula for solving a quadratic equation. We have also discussed some common mistakes to avoid when solving quadratic equations and some real-world applications of quadratic equations.