Problem 21 - Precalculus: Mathematical for Calculus

Question

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

Step-by-Step Solution

Answer

Intersection points: \((0, 0)\), \((1, -1)\), \((-2, -4)\).

Step 1: Understand the Problem

We have two equations: \( x^2 + y = 0 \) and \( x^3 - 2x - y = 0 \). We need to solve the system of these two equations to find the intersection point(s) of their graphs.

Step 2: Solve the First Equation for y

From the first equation, we can solve for \( y \). We have \( x^2 + y = 0 \). Solving for \( y \), we get: \( y = -x^2 \).

Step 3: Substitute y in the Second Equation

Substitute \( y = -x^2 \) into the second equation \( x^3 - 2x - y = 0 \). This gives us: \( x^3 - 2x - (-x^2) = 0 \).

Step 4: Simplify the Equation

The equation from the substitution becomes \( x^3 - 2x + x^2 = 0 \). Rearrange the terms: \( x^3 + x^2 - 2x = 0 \).

Step 5: Factor the Equation

Factor out the common term \( x \): \( x(x^2 + x - 2) = 0 \). The solutions of this equation occur when \( x = 0 \) or when the quadratic \( x^2 + x - 2 = 0 \) is zero.

Step 6: Solve the Quadratic Equation

Solve the quadratic equation \( x^2 + x - 2 = 0 \) using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 1 \), \( c = -2 \). Substituting these values, we get: \( x = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} \). The solutions are \( x = 1 \) and \( x = -2 \).

Step 7: Find Corresponding y-values

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

Step 8: Conclusion

The intersection points of the graphs of the equations are \((0, 0)\), \((1, -1)\), and \((-2, -4)\).

Key Concepts

Intersection Points

When we talk about intersection points in a system of equations, we mean places where the graphs of these equations meet. For the given system of two equations \( x^2 + y = 0 \) and \( x^3 - 2x - y = 0 \), finding their intersection points involves solving the equations simultaneously. This means finding the values of \( x \) and \( y \) that satisfy both equations at the same time.

• Intersection points provide real solutions that belong to both curves.
• They help us understand how different functions relate on the same coordinate plane.

By solving the given system, we've found three intersection points: \((0, 0)\), \((1, -1)\), and \((-2, -4)\). These are the points where the two graphs cross each other on a coordinate plane.

Quadratic Equation

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients with \( a eq 0 \). In our problem, part of our process involved solving the quadratic equation \( x^2 + x - 2 = 0 \).

• Quadratic equations often have two solutions because they are second-degree polynomials, represented by the highest exponent of 2.
• These solutions may be real or complex numbers, depending on the discriminant \( b^2 - 4ac \).

To solve \( x^2 + x - 2 = 0 \), we used the quadratic formula, which is a reliable tool for finding the roots of any quadratic equation: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]By applying the formula with \( a = 1 \), \( b = 1 \), and \( c = -2 \), we obtained the solutions \( x = 1 \) and \( x = -2 \).

Factoring Polynomials

Factoring polynomials is an essential skill in solving higher degree polynomials. It involves expressing a polynomial as a product of its factors. In the equation \( x^3 + x^2 - 2x = 0 \), we simplified our work by factoring out a common factor, which in this case is \( x \).

• Start by identifying the greatest common factor (GCF) of all terms. Here, the GCF is \( x \).
• Rewrite the polynomial as a product: \( x(x^2 + x - 2) = 0 \).

Once the polynomial was factored, it became much easier to solve because we could then set each factor equal to zero and solve for \( x \). This method significantly reduces the complexity of polynomial equations, breaking them down into manageable parts.

Substitution Method

The substitution method is a powerful technique used in solving systems of equations. It involves replacing one variable with an expression involving the other variable from one equation into the other equation. Here’s how we applied it in the given system:

• From the first equation, \( x^2 + y = 0 \), we solved for \( y \), giving us \( y = -x^2 \).
• We substituted \( y = -x^2 \) into the second equation \( x^3 - 2x - y = 0 \).

This substitution changed the original system into a single equation with one variable: \( x^3 + x^2 - 2x = 0 \). This simplification allowed us to focus on solving for \( x \) first. After finding the values of \( x \), it became straightforward to substitute back to find the corresponding \( y \) values. By using substitution, we made solving the system far simpler and more direct.