Problem 22 - Precalculus: Mathematical for Calculus

Question

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

Step-by-Step Solution

Answer

The intersection points are (0, 0) and (3, ±√3).

Step 1: Understand the System of Equations

The given system of equations is \(x^2 + y^2 = 4x\) and \(x = y^2\). The first equation represents a circle, while the second represents a parabola. We need to find the points where these two graphs intersect.

Step 2: Substitute to Simplify

Substitute the expression for \(x\) from the second equation \(x = y^2\) into the first equation \(x^2 + y^2 = 4x\). This results in \((y^2)^2 + y^2 = 4(y^2)\).

Step 3: Simplify the New Equation

Simplify \((y^2)^2 + y^2 = 4(y^2)\) to get \(y^4 + y^2 = 4y^2\), then rearrange it to \(y^4 - 3y^2 = 0\).

Step 4: Factor the Equation

Factor the equation \(y^4 - 3y^2 = 0\) as \(y^2(y^2 - 3) = 0\). This gives us two factors: \(y^2 = 0\) and \(y^2 - 3 = 0\).

Step 5: Solve for y

Solve \(y^2 = 0\) to get \(y = 0\). Solve \(y^2 - 3 = 0\) to obtain \(y = \pm\sqrt{3}\). These are the \(y\)-values of the intersection points.

Step 6: Solve for x

Use the expression \(x = y^2\) to find the \(x\)-values corresponding to each \(y\). For \(y = 0\), \(x = 0^2 = 0\). For \(y = \pm\sqrt{3}\), \(x = (\pm\sqrt{3})^2 = 3\).

Step 7: Identify Intersection Points

The coordinates of the intersection points are \((0, 0)\) and \((3, \pm\sqrt{3})\). These are the points where the graphs of the equations intersect.

Key Concepts

System of Equations

A system of equations involves finding values for variables that satisfy all equations involved. In this exercise, we have two equations: a circle equation and a parabola equation. Our goal is to identify where these two graphs meet, known as intersection points. To find these points, we combine or substitute one equation into another, enabling us to solve for the variables effectively.

When dealing with a system like this, remember:

• Identify each equation and the type of graph it represents (e.g., circle, parabola).
• Look for ways to substitute or combine equations to simplify the process of finding variable values.
• Solve the simplified equations to find possible solutions for both variables, typically denoted as \(x\) and \(y\).

Finding the intersection points is much like solving a puzzle, where each piece of information provided by an equation leads you closer to the complete solution.

Circle Graph

A circle graph is a visual representation of a circle in the coordinate plane, expressed by an equation. In this exercise, the circle is represented by the equation \(x^2 + y^2 = 4x\). To simplify this and better understand its nature, we can complete the square:

Firstly, rearrange and complete the square for the term involving \(x\): \(x^2 - 4x + y^2 = 0\).
To complete the square, think of \(x^2 - 4x\) as \((x - 2)^2 - 4\) by adding and subtracting 4. This reorganization lets us rewrite the equation as \( (x - 2)^2 + y^2 = 4 \), a standard circle form centered at (2,0) with a radius of 2.

Key points of understanding a circle graph:

• The general form is \((x - h)^2 + (y - k)^2 = r^2\), where (h, k) is the center and \(r\) is the radius.
• Circle graphs are symmetric and represent all points equidistant from the center.

In the context of the system, finding where this circle meets a parabola graph becomes critical to solving the exercise.

Parabola Graph

A parabola graph represents a curve, generated from quadratic equations, with a distinct U-shape, either opening up or down. In our exercise, the parabola is given by the equation \(x = y^2\).

This equation is already set in a "sideways" format which is typical for a parabola expressed with \(x\) in terms of \(y\):

• Standard form is typically \(y = ax^2 + bx + c\) for vertical parabolas; our parabola being horizontal is expressed as \(x = ay^2 + by + c\).

The direction of opening is determined by the coefficient of \(y^2\). In this case, the positive term implies it opens to the right.

Key features of parabola graphs include:

• The vertex, which is the highest or lowest point. Here it would naturally sit at (0,0) assuming no additional transformations.
• Symmetry along the directrix or axis of the parabola.

Understanding how this graph interacts with other graph types, like circles, helps us pinpoint their intersections, which are crucial in finding the solution to this system of equations.