Graph The Equation Y = − X 2 + 12 X − 35 Y = -x^2 + 12x - 35 Y = − X 2 + 12 X − 35 , Including The Roots And The Vertex.Click To Plot Points. Click Points To Delete Them.

by ADMIN 171 views

Introduction

Graphing a quadratic equation is an essential skill in mathematics, and it can be used to model various real-world situations. In this article, we will focus on graphing the equation y=−x2+12x−35y = -x^2 + 12x - 35, including the roots and the vertex. We will use the concept of factoring, completing the square, and the properties of quadratic functions to graph the equation.

Understanding the Equation

The given equation is a quadratic equation in the form of y=ax2+bx+cy = ax^2 + bx + c, where a=−1a = -1, b=12b = 12, and c=−35c = -35. The coefficient of the squared term, aa, is negative, which means that the parabola will open downwards.

Finding the Roots

To find the roots of the equation, we need to factorize the quadratic expression. We can start by finding two numbers whose product is equal to the product of the coefficient of the squared term and the constant term, and whose sum is equal to the coefficient of the linear term.

import sympy as sp

x = sp.symbols('x')



equation = -x**2 + 12*x - 35



roots = sp.solve(equation, x)


print(roots)

The output of the above code will be [5, 7], which are the roots of the equation.

Finding the Vertex

The vertex of a parabola is the point where the parabola changes direction. To find the vertex, we can use the formula x = -b / 2a, where a and b are the coefficients of the squared term and the linear term, respectively.

# Define the coefficients
a = -1
b = 12


x_vertex = -b / (2 * a)


print(x_vertex)

The output of the above code will be 6, which is the x-coordinate of the vertex.

Graphing the Equation

Now that we have found the roots and the vertex, we can graph the equation. We can use a graphing calculator or a computer program to graph the equation.

import matplotlib.pyplot as plt
import numpy as np


x_values = np.linspace(-10, 10, 400)



y_values = -x_values**2 + 12*x_values - 35



plt.plot(x_values, y_values)
plt.scatter([5, 7], [0, 0], color='red')
plt.scatter([6], [-1], color='green')
plt.xlabel('x')
plt.ylabel('y')
plt.title('Graph of the Equation y=−x2+12x−35y = -x^2 + 12x - 35')
plt.grid(True)
plt.axhline(0, color='black')
plt.axvline(0, color='black')
plt.show()

The above code will create a graph of the equation, including the roots and the vertex.

Conclusion

In this article, we have graphed the equation y=−x2+12x−35y = -x^2 + 12x - 35, including the roots and the vertex. We have used the concept of factoring, completing the square, and the properties of quadratic functions to graph the equation. We have also used a graphing calculator or a computer program to graph the equation.

References

Further Reading

Q: What is the equation y=−x2+12x−35y = -x^2 + 12x - 35?

A: The equation y=−x2+12x−35y = -x^2 + 12x - 35 is a quadratic equation in the form of y=ax2+bx+cy = ax^2 + bx + c, where a=−1a = -1, b=12b = 12, and c=−35c = -35. This equation represents a parabola that opens downwards.

Q: How do I find the roots of the equation?

A: To find the roots of the equation, you can factorize the quadratic expression. You can start by finding two numbers whose product is equal to the product of the coefficient of the squared term and the constant term, and whose sum is equal to the coefficient of the linear term. In this case, the roots are x=5x = 5 and x=7x = 7.

Q: How do I find the vertex of the parabola?

A: To find the vertex of the parabola, you can use the formula x=−b/2ax = -b / 2a, where aa and bb are the coefficients of the squared term and the linear term, respectively. In this case, the x-coordinate of the vertex is x=6x = 6.

Q: How do I graph the equation?

A: You can graph the equation using a graphing calculator or a computer program. You can also use the concept of factoring, completing the square, and the properties of quadratic functions to graph the equation.

Q: What is the significance of the roots and the vertex of the parabola?

A: The roots of the parabola represent the points where the parabola intersects the x-axis. The vertex of the parabola represents the point where the parabola changes direction.

Q: Can I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model real-world situations?

A: Yes, you can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model real-world situations. For example, you can use this equation to model the height of a projectile as a function of time.

Q: How do I determine the direction of the parabola?

A: You can determine the direction of the parabola by looking at the coefficient of the squared term. If the coefficient is positive, the parabola opens upwards. If the coefficient is negative, the parabola opens downwards.

Q: Can I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in other areas of mathematics?

A: Yes, you can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in other areas of mathematics. For example, you can use this equation to solve problems in algebra, geometry, and calculus.

Q: How do I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in real-world situations?

A: You can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in real-world situations by substituting the given values into the equation and solving for the unknown variable.

Q: Can I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of an object?

A: Yes, you can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of an object. For example, you can use this equation to model the height of a projectile as a function of time.

Q: How do I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in physics?

A: You can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in physics by substituting the given values into the equation and solving for the unknown variable.

Q: Can I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of a particle?

A: Yes, you can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of a particle. For example, you can use this equation to model the position of a particle as a function of time.

Q: How do I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in engineering?

A: You can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in engineering by substituting the given values into the equation and solving for the unknown variable.

Q: Can I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of a system?

A: Yes, you can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of a system. For example, you can use this equation to model the position of a system as a function of time.

Q: How do I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in computer science?

A: You can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in computer science by substituting the given values into the equation and solving for the unknown variable.

Q: Can I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of a robot?

A: Yes, you can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of a robot. For example, you can use this equation to model the position of a robot as a function of time.

Q: How do I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in robotics?

A: You can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in robotics by substituting the given values into the equation and solving for the unknown variable.

Q: Can I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of a vehicle?

A: Yes, you can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of a vehicle. For example, you can use this equation to model the position of a vehicle as a function of time.

Q: How do I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in transportation?

A: You can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in transportation by substituting the given values into the equation and solving for the unknown variable.

Q: Can I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of a person?

A: Yes, you can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of a person. For example, you can use this equation to model the position of a person as a function of time.

Q: How do I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in human motion?

A: You can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in human motion by substituting the given values into the equation and solving for the unknown variable.

Q: Can I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of an animal?

A: Yes, you can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of an animal. For example, you can use this equation to model the position of an animal as a function of time.

Q: How do I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in animal motion?

A: You can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in animal motion by substituting the given values into the equation and solving for the unknown variable.

Q: Can I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of a system of particles?

A: Yes, you can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to model the motion of a system of particles. For example, you can use this equation to model the position of a system of particles as a function of time.

Q: How do I use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in systems of particles?

A: You can use the equation y=−x2+12x−35y = -x^2 + 12x - 35 to solve problems in systems of particles by substituting the given values into the equation