The Zeros Of A Parabola Are -5 And -3. The Point { (0,60)$}$ Is On The Graph As Represented By The Equation:${ 60 = A(0+5)(0+3) }$What Is The Value Of { A$}$?A. { -7.5$}$ B. { -4$}$ C.

Understanding the Problem

The problem presents a parabola with zeros at -5 and -3, and a point (0,60) on its graph. The equation representing this parabola is given as $60 = a(0+5)(0+3)$. Our goal is to find the value of 'a', which is a crucial parameter in the equation of a parabola.

The Equation of a Parabola

The general equation of a parabola can be written as $y = a(x - r)(x - s)$, where 'a' is the coefficient, and 'r' and 's' are the zeros of the parabola. In this case, the zeros are given as -5 and -3, so the equation can be written as $y = a(x + 5)(x + 3)$.

Substituting the Given Point

We are given that the point (0,60) lies on the graph of the parabola. Substituting this point into the equation, we get $60 = a(0 + 5)(0 + 3)$. This equation can be simplified to $60 = a(5)(3)$.

Solving for 'a'

To find the value of 'a', we need to isolate 'a' in the equation. Dividing both sides of the equation by 15 (which is the product of 5 and 3), we get $a = \frac{60}{15}$.

Calculating the Value of 'a'

Now, we can calculate the value of 'a' by simplifying the fraction. $a = \frac{60}{15} = 4$.

Conclusion

In conclusion, the value of 'a' in the equation $60 = a(0+5)(0+3)$ is 4. This means that the parabola has a coefficient of 4, and its zeros are at -5 and -3.

Discussion and Analysis

The problem presented in this article is a classic example of a quadratic equation. The equation of a parabola can be written in the form $y = a(x - r)(x - s)$, where 'a' is the coefficient, and 'r' and 's' are the zeros of the parabola. In this case, the zeros are given as -5 and -3, and the point (0,60) lies on the graph of the parabola.

Importance of Quadratic Equations

Quadratic equations are an essential part of mathematics, and they have numerous applications in various fields such as physics, engineering, and economics. The equation of a parabola can be used to model real-world situations such as the trajectory of a projectile, the motion of a pendulum, and the growth of a population.

Solving Quadratic Equations

Solving quadratic equations involves finding the values of the zeros of the parabola. There are several methods to solve quadratic equations, including factoring, the quadratic formula, and graphing. In this article, we used the method of substitution to solve the equation.

Real-World Applications

Quadratic equations have numerous real-world applications. For example, in physics, the equation of a parabola can be used to model the trajectory of a projectile. In engineering, quadratic equations can be used to design the shape of a bridge or a building. In economics, quadratic equations can be used to model the growth of a population or the demand for a product.

Conclusion

In conclusion, the value of 'a' in the equation $60 = a(0+5)(0+3)$ is 4. This means that the parabola has a coefficient of 4, and its zeros are at -5 and -3. Quadratic equations are an essential part of mathematics, and they have numerous applications in various fields. Solving quadratic equations involves finding the values of the zeros of the parabola, and there are several methods to solve quadratic equations.

Final Answer

The final answer is $\boxed{4}$.

Understanding the Problem

The problem presented in this article is a classic example of a quadratic equation. The equation of a parabola can be written in the form $y = a(x - r)(x - s)$, where 'a' is the coefficient, and 'r' and 's' are the zeros of the parabola. In this case, the zeros are given as -5 and -3, and the point (0,60) lies on the graph of the parabola.

Q&A

Q: What is the general equation of a parabola?

A: The general equation of a parabola can be written as $y = a(x - r)(x - s)$, where 'a' is the coefficient, and 'r' and 's' are the zeros of the parabola.

Q: What are the zeros of the parabola in this problem?

A: The zeros of the parabola in this problem are -5 and -3.

Q: How do we find the value of 'a' in the equation?

A: To find the value of 'a', we need to isolate 'a' in the equation. We can do this by substituting the given point (0,60) into the equation and solving for 'a'.

Q: What is the value of 'a' in the equation $60 = a(0+5)(0+3)$?

A: The value of 'a' in the equation $60 = a(0+5)(0+3)$ is 4.

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

A: Quadratic equations have numerous real-world applications. For example, in physics, the equation of a parabola can be used to model the trajectory of a projectile. In engineering, quadratic equations can be used to design the shape of a bridge or a building. In economics, quadratic equations can be used to model the growth of a population or the demand for a product.

Q: How do we solve quadratic equations?

A: There are several methods to solve quadratic equations, including factoring, the quadratic formula, and graphing. In this article, we used the method of substitution to solve the equation.

Q: What is the importance of quadratic equations in mathematics?

A: Quadratic equations are an essential part of mathematics, and they have numerous applications in various fields. The equation of a parabola can be used to model real-world situations such as the trajectory of a projectile, the motion of a pendulum, and the growth of a population.

Conclusion

In conclusion, the value of 'a' in the equation $60 = a(0+5)(0+3)$ is 4. This means that the parabola has a coefficient of 4, and its zeros are at -5 and -3. Quadratic equations are an essential part of mathematics, and they have numerous applications in various fields. Solving quadratic equations involves finding the values of the zeros of the parabola, and there are several methods to solve quadratic equations.

Final Answer

The final answer is $\boxed{4}$.

Additional Resources

For more information on quadratic equations and their applications, please refer to the following resources:

• Textbooks: "Algebra and Trigonometry" by Michael Sullivan, "Calculus" by James Stewart
• Online Resources: Khan Academy, MIT OpenCourseWare, Wolfram Alpha
• Software: Mathematica, Maple, MATLAB

Practice Problems

Try solving the following quadratic equations:

• $x^2 + 5x + 6 = 0$
• $x^2 - 4x + 4 = 0$
• $x^2 + 2x - 15 = 0$

Use the methods discussed in this article to solve these equations and find the values of the zeros.