Problem 17 - Algebra and Trigonometry

Question

\(15-18\) . Two equations and their graphs are given. Find the inter- section point(s) of the graphs by solving the system. $$ \left\\{\begin{aligned} x^{2}+y &=0 \\ x^{3}-2 x-y &=0 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The intersection points are (0, 0), (1, -1), and (-2, -4).

Step 1: Understand the Given System of Equations

The system consists of two equations:1. \( x^2 + y = 0 \)2. \( x^3 - 2x - y = 0 \).The task is to find the intersection points of the graphs represented by these equations by solving the system.

Step 2: Express y in Terms of x from the First Equation

From the first equation, \( x^2 + y = 0 \), express \( y \) in terms of \( x \):\[ y = -x^2 \].

Step 3: Substitute y in the Second Equation

Substitute \( y = -x^2 \) into the second equation \( x^3 - 2x - y = 0 \):\[ x^3 - 2x - (-x^2) = 0 \].

Step 4: Simplify and Solve the Equation

The substitution gives:\[ x^3 - 2x + x^2 = 0 \]Reorder the terms:\[ x^3 + x^2 - 2x = 0 \]Factor out \( x \):\[ x(x^2 + x - 2) = 0 \].This implies \( x = 0 \) or \( x^2 + x - 2 = 0 \).

Step 5: Solve the Quadratic Equation

Solve the quadratic equation \( x^2 + x - 2 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1, c = -2 \):\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \]\[ x = \frac{-1 \pm \sqrt{1 + 8}}{2} \]\[ x = \frac{-1 \pm 3}{2} \].Thus, \( x = 1 \) or \( x = -2 \).

Step 6: Find Corresponding y Values

For each value of \( x \), find \( y \) using \( y = -x^2 \):- If \( x = 0 \), \( y = -0^2 = 0 \). Thus the point is \((0, 0)\).- If \( x = 1 \), \( y = -1^2 = -1 \). Thus the point is \((1, -1)\).- If \( x = -2 \), \( y = -(-2)^2 = -4 \). Thus the point is \((-2, -4)\).

Step 7: List the Intersection Points

The intersection points are: \((0, 0), (1, -1), (-2, -4)\).

Key Concepts

Intersection Points

When we talk about intersection points in the context of equations, we're referring to the places where the graphs of these equations cross each other. Imagine drawing two curves or lines on a graph. Wherever they touch or overlap is called an intersection point. These points are significant because they represent solutions that satisfy both equations at the same time.
To find intersection points, we solve the system of equations given to us. In this exercise, our system consists of two equations that describe a parabola and a more complex curve. By solving these equations, we find the specific x and y values that make both equations true.
This process usually involves finding common variables or shared values between the equations, leading us to the intersection points where the graphs of these equations overlap. In our current problem, the solutions are points

• (0, 0)
• (1, -1)
• (-2, -4)

These points indicate where on the coordinate plane both graphs meet.

Quadratic Equation

A quadratic equation is a polynomial equation of degree 2, generally expressed in the form \( ax^2 + bx + c = 0 \). In our problem, once we substituted the expression for \( y \) from the first equation into the second equation, we ended up with a quadratic equation: \( x^2 + x - 2 = 0 \).
Quadratic equations are fascinating because they create parabolas when graphed. Parabolas are symmetrical, U-shaped curves. The solutions to these equations are the values of \( x \) where the parabola crosses the x-axis. These solutions can be found using various methods, such as factoring, completing the square, or using the quadratic formula.
In this example, we used the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1 \), \( b = 1 \), and \( c = -2 \). Applying this formula gives us the x-values \( 1 \) and \( -2 \), which are part of our solution for the intersection points.

Substitution Method

The substitution method is a technique used to solve a system of equations. It involves solving one equation for one variable and then substituting that expression into the other equation. This reduces the two-variable system to a single equation with one variable, making it easier to solve.
In our exercise, we first solved the equation \( x^2 + y = 0 \) to express \( y \) in terms of \( x \): \( y = -x^2 \). We then substituted \( -x^2 \) for \( y \) in the second equation \( x^3 - 2x - y = 0 \). This substitution reformed the equation into:
\[x^3 - 2x + x^2 = 0\]
This process simplifies the system by reducing it to a problem that involves solving a quadratic equation, which is straightforward. The substitution method not only simplifies systems of equations but also highlights relationships between variables that can help in solving more complex problems.