Determining the Maximum and Minimum Values

Oct 13, 2025 by magicalceylontours.com 23 views

Introduction

Graphs are a powerful tool in mathematics, allowing us to visualize and analyze complex relationships between variables. In this article, we will explore how to use graphs to solve three different problems related to the equation $y = x^2 + 3x - 2$ . We will first plot the equation, then find the value of $x$ when $y = 6$ , and finally determine the maximum and minimum values of the equation.

Plotting the Equation

To plot the equation $y = x^2 + 3x - 2$ , we need to find the values of $y$ for a range of values of $x$ . We are given that $x$ ranges from -5 to 2.2. We can use a graphing calculator or a computer program to plot the equation.

import numpy as np
import matplotlib.pyplot as plt

# Define the function
def f(x):
    return x**2 + 3*x - 2

# Generate x values
x = np.linspace(-5, 2.2, 400)

# Calculate y values
y = f(x)

# Plot the graph
plt.plot(x, y)
plt.title('Graph of y = x^2 + 3x - 2')
plt.xlabel('x')
plt.ylabel('y')
plt.grid(True)
plt.axhline(0, color='black')
plt.axvline(0, color='black')
plt.show()

The resulting graph shows a parabola that opens upwards, with a minimum point at $x = -1.5$ and $y = -3.5$ . The graph also shows that the equation has a maximum value at $x = 2.2$ and $y = 12.2$ .

Finding the Value of x When y = 6

To find the value of $x$ when $y = 6$ , we need to set $y = 6$ in the equation and solve for $x$ . We can do this by rearranging the equation to get:

x^2 + 3x - 2 = 6

Subtracting 6 from both sides gives:

x^2 + 3x - 8 = 0

We can solve this quadratic equation using the quadratic formula:

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

In this case, $a = 1$ , $b = 3$ , and $c = -8$ . Plugging these values into the formula gives:

x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-8)}}{2(1)}

Simplifying the expression gives:

x = \frac{-3 \pm \sqrt{37}}{2}

Since $x$ must be a real number, we take the positive square root:

x = \frac{-3 + \sqrt{37}}{2}

Determining the Maximum and Minimum Values

To determine the maximum and minimum values of the equation, we need to find the critical points of the function. The critical points occur when the derivative of the function is equal to zero or undefined.

The derivative of the function is:

f'(x) = 2x + 3

Setting the derivative equal to zero gives:

2x + 3 = 0

Solving for $x$ gives:

x = -\frac{3}{2}

This is the critical point of the function. To determine whether this point corresponds to a maximum or minimum, we need to examine the behavior of the function near this point.

Since the function is a parabola that opens upwards, the critical point corresponds to a minimum value. The minimum value of the function occurs at $x = -1.5$ and $y = -3.5$ .

To determine the maximum value of the function, we need to examine the behavior of the function as $x$ approaches infinity. Since the function is a parabola that opens upwards, the function approaches infinity as $x$ approaches infinity.

Therefore, the maximum value of the function occurs at $x = 2.2$ and $y = 12.2$ .

Conclusion

In this article, we have used graphs to solve three different problems related to the equation $y = x^2 + 3x - 2$ . We have plotted the equation, found the value of $x$ when $y = 6$ , and determined the maximum and minimum values of the equation. We have shown that the equation has a minimum point at $x = -1.5$ and $y = -3.5$ , and a maximum value at $x = 2.2$ and $y = 12.2$ .

References

[1] "Graphing Calculators" by Math Is Fun
[2] "Quadratic Formula" by Math Is Fun
[3] "Derivatives" by Math Is Fun

Appendix

The following is a list of the equations and formulas used in this article:

$y = x^2 + 3x - 2$
$x^2 + 3x - 8 = 0$
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$f'(x) = 2x + 3$
$x = -\frac{3}{2}$
Solving Problems Using Graphs: A Mathematical Approach - Q&A ===========================================================

Introduction

In our previous article, we explored how to use graphs to solve three different problems related to the equation $y = x^2 + 3x - 2$ . We plotted the equation, found the value of $x$ when $y = 6$ , and determined the maximum and minimum values of the equation. In this article, we will answer some frequently asked questions related to the topic.

Q&A

Q: What is the purpose of plotting the equation?

A: Plotting the equation helps us visualize the relationship between the variables $x$ and $y$ . It allows us to see the shape of the graph and identify any patterns or trends.

Q: How do I determine the maximum and minimum values of the equation?

A: To determine the maximum and minimum values of the equation, we need to find the critical points of the function. The critical points occur when the derivative of the function is equal to zero or undefined. We can then examine the behavior of the function near these points to determine whether they correspond to a maximum or minimum value.

Q: What is the difference between a maximum and a minimum value?

A: A maximum value is the largest value that the function can take, while a minimum value is the smallest value that the function can take. In the case of the equation $y = x^2 + 3x - 2$ , the maximum value occurs at $x = 2.2$ and $y = 12.2$ , while the minimum value occurs at $x = -1.5$ and $y = -3.5$ .

Q: How do I find the value of $x$ when $y = 6$ ?

A: To find the value of $x$ when $y = 6$ , we need to set $y = 6$ in the equation and solve for $x$ . We can do this by rearranging the equation to get:

x^2 + 3x - 8 = 0

We can then solve this quadratic equation using the quadratic formula:

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

In this case, $a = 1$ , $b = 3$ , and $c = -8$ . Plugging these values into the formula gives:

x = \frac{-3 \pm \sqrt{37}}{2}

Since $x$ must be a real number, we take the positive square root:

x = \frac{-3 + \sqrt{37}}{2}

Q: What is the significance of the critical points?

A: The critical points are significant because they correspond to the maximum and minimum values of the function. In the case of the equation $y = x^2 + 3x - 2$ , the critical point occurs at $x = -1.5$ and corresponds to the minimum value of the function.

Q: How do I determine the behavior of the function near the critical points?

A: To determine the behavior of the function near the critical points, we need to examine the derivative of the function. If the derivative is positive, the function is increasing near the critical point. If the derivative is negative, the function is decreasing near the critical point.

Q: What is the relationship between the derivative and the critical points?

A: The derivative of the function is related to the critical points in that it determines the behavior of the function near the critical points. If the derivative is zero or undefined at a critical point, the function has a maximum or minimum value at that point.

Conclusion

In this article, we have answered some frequently asked questions related to the topic of solving problems using graphs. We have discussed the purpose of plotting the equation, how to determine the maximum and minimum values of the equation, and the significance of the critical points. We have also examined the relationship between the derivative and the critical points.

References

[1] "Graphing Calculators" by Math Is Fun
[2] "Quadratic Formula" by Math Is Fun
[3] "Derivatives" by Math Is Fun

Appendix

The following is a list of the equations and formulas used in this article:

$y = x^2 + 3x - 2$
$x^2 + 3x - 8 = 0$
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$f'(x) = 2x + 3$
$x = -\frac{3}{2}$