Problem 20 - Precalculus: Mathematical for Calculus

Question

Two equations and their graphs are given. Find the intersection point(s) of the graphs by solving the system. $$\left\\{\begin{array}{l}x-y^{2}=-4 \\\x-y=2\end{array}\right.$$ GRAPH CANT COPY

Step-by-Step Solution

Answer

The intersections are at (5, 3) and (0, -2).

Step 1: Understand the System of Equations

We have the system of equations \(x - y^2 = -4\) and \(x - y = 2\). We aim to find a value of \(x\) and \(y\) that satisfies both equations together.

Step 2: Solve One Equation for x or y

From the second equation \(x - y = 2\), we can easily express \(x\) in terms of \(y\): \[x = y + 2\]

Step 3: Substitute into the First Equation

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

Step 4: Rearrange and Simplify

Rearrange the equation \((y + 2) - y^2 = -4\) by combining like terms. This gives: \[y + 2 - y^2 = -4\]\[-y^2 + y + 2 = -4\]Then, add 4 to both sides: \[-y^2 + y + 6 = 0\]

Step 5: Solve the Quadratic Equation

Rearrange \(-y^2 + y + 6 = 0\) into standard quadratic form: \[y^2 - y - 6 = 0\]Use the quadratic formula to solve for \(y\): \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Where \(a = 1\), \(b = -1\), and \(c = -6\): \[y = \frac{1 \pm \sqrt{1 + 24}}{2}\]\[y = \frac{1 \pm \sqrt{25}}{2}\]\[y = \frac{1 \pm 5}{2}\]Therefore, \(y = 3\) and \(y = -2\).

Step 6: Find Corresponding x Values

Substitute each value of \(y\) back into \(x = y + 2\):For \(y = 3\):\[x = 3 + 2 = 5\]For \(y = -2\):\[x = -2 + 2 = 0\]

Step 7: Write the Intersection Points

The intersection points are where both \(x\) and \(y\) satisfy both equations:The points are \((x = 5, y = 3)\) and \((x = 0, y = -2)\).

Key Concepts

Intersection Points

Intersection points are a crucial concept when working with systems of equations. They represent the coordinates where the graphs of these equations meet on a coordinate plane.

In simpler terms, if you draw each equation as a line or curve on a graph, the intersection points are where they cross each other. These points are the solutions to the system of equations, meaning they satisfy both equations simultaneously.

If you visualize this with our given equations, you can imagine graphing them helps locate these meeting places visually.

• The equation \(x - y^2 = -4\) is in a quadratic form, hinting at a parabola shape.
• The equation \(x - y = 2\) forms a straight line.

By solving the system algebraically, you pinpoint exactly where these two shapes intersect. From our solution, the intersection points are \((5, 3)\) and \((0, -2)\). These are the values of \(x\) and \(y\) that satisfy both equations.

Quadratic Formula

The quadratic formula is a powerful tool for solving polynomial equations of the form \(ax^2 + bx + c = 0\). It's especially useful when factoring the equation seems tricky or impossible.

The formula is:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
where \(a\), \(b\), and \(c\) are coefficients from the quadratic equation.

For our exercise, once we manipulated the equation into the standard quadratic form \(y^2 - y - 6 = 0\), we applied this formula. Here's a breakdown of our steps:

• Identify \(a = 1\), \(b = -1\), \(c = -6\).
• Calculate the discriminant \(b^2 - 4ac = 1 + 24 = 25\).
• The square root of 25 is 5, simplifying the formula further.
• Compute solutions for \(y\): \(\frac{1 \pm 5}{2}\), giving \(y = 3\) and \(y = -2\).

The quadratic formula efficiently finds these precise values of \(y\), leading us to discover our intersection points when paired with the corresponding \(x\)-values found through substitution.

Graphing

Graphing is a visual way to understand the relationships between algebraic equations and their solutions. By drawing the graphs of equations, you can quickly see where they intersect or meet, revealing solutions to the system.

The process involves plotting each equation on a coordinate grid:

• For the equation \(x - y^2 = -4\), graphing involves creating a curved shape or parabola.
• For the equation \(x - y = 2\), graphing results in a straight line.

The point where the line intersects the parabola on the graph corresponds to the intersection points. These are exactly the solutions we found algebraically.

Graphing not only helps you visually check your solution but also provides an intuitive understanding of how each function behaves and interacts with others. It reinforces the concept that solutions to a system of equations are, in essence, the meeting points of the graphs representing those equations.