Problem 16 - Precalculus Mathematics for Calculus

Question

Two equations and their graphs are given. Find the intersection point(s) of the graphs by solving the system. $$\left\\{\begin{array}{l} x-y^{2}=-4 \\ x-y=2 \end{array}\right.$$ (GRAPH CAN'T COPY).

Step-by-Step Solution

Answer

Intersection points are (5, 3) and (0, -2).

Step 1: Identify the System of Equations

We have two equations given in the system:1. \( x - y^2 = -4 \)2. \( x - y = 2 \)Our task is to find the point(s) of intersection of the graphs by solving this system of equations.

Step 2: Solve the Second Equation for x

From the second equation, we solve for \( x \):\[ x - y = 2 \]Add \( y \) to both sides to isolate \( x \):\[ x = y + 2 \]

Step 3: Substitute x in the First Equation

Now that we have \( x = y + 2 \), substitute \( x \) in the first equation:\[ (y + 2) - y^2 = -4 \]

Step 4: Rearrange the Equation

Rearrange the equation by simplifying and moving all terms to one side:\[ y + 2 - y^2 = -4 \]\[ -y^2 + y + 2 + 4 = 0 \]\[ -y^2 + y + 6 = 0 \]

Step 5: Solve the Quadratic Equation

Rewrite the quadratic equation \( -y^2 + y + 6 = 0 \) as \( y^2 - y - 6 = 0 \) by multiplying through by -1. Now solve using the quadratic formula:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = 1, b = -1, c = -6 \).\[ y = \frac{1 \pm \sqrt{1 + 24}}{2} \]\[ y = \frac{1 \pm \sqrt{25}}{2} \]\[ y = \frac{1 \pm 5}{2} \]So, \( y = 3 \) or \( y = -2 \).

Step 6: Find Corresponding x-values

For each \( y \) value, use \( x = y + 2 \) to find the corresponding \( x \) value:- If \( y = 3 \):\( x = 3 + 2 = 5 \)- If \( y = -2 \):\( x = -2 + 2 = 0 \)

Step 7: Determine the Intersection Points

The solutions correspond to intersection points where both equations are satisfied:- The first intersection point is \((5, 3)\).- The second intersection point is \((0, -2)\).

Key Concepts

Intersection Points

Intersection points are the points where two graphs meet on a coordinate plane. These points are important because they satisfy both equations simultaneously. When you solve a system of equations, you are essentially looking for these intersection points. In this particular problem, we found two intersection points: \((5, 3)\) and \((0, -2)\). These are the solutions to the system of equations because they lie on the graphs of both equations. Finding intersection points involves solving the equations together, which can often be done by substitution or elimination methods. In graphical terms, if you plot both equations, the intersection points are where the curves cross each other on the graph. These results show that even nonlinear systems, such as one involving quadratic equations, have distinct intersection points that we can calculate.

Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). They are called quadratic because the highest power of the variable \( x \) is \( 2 \). In this exercise, the quadratic equation we worked with was \( y^2 - y - 6 = 0 \), which arose from manipulating the given system. Quadratic equations can be solved using several methods such as factoring, completing the square, or using the quadratic formula. In our solution, we used the quadratic formula:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -1 \), and \( c = -6 \). By substituting these values, we found the solutions for \( y \), which ultimately helped us find the intersection points for the system.

Solving Equations

Solving equations is a fundamental part of mathematics, requiring methods like substitution and manipulation to find variable values that satisfy all given equations. Here, we were given a system of equations involving both a linear and a quadratic equation:- \( x - y^2 = -4 \)- \( x - y = 2 \)To solve these equations, we first simplified one of them. Solving for \( x \) in the simpler linear equation gave us \( x = y + 2 \). We then substituted \( x = y + 2 \) into the quadratic equation to eliminate \( x \) from that equation, leaving us with an equation in terms of \( y \) alone.This manipulation led to a quadratic equation, which we solved using the quadratic formula. By finding \( y \) values first, and then using them to find corresponding \( x \) values, we resolved the entire system and identified the intersection points.