Problem 43 - Precalculus with Limits

Question

Solving a System of Equations Graphically In Exercises \(33-44,\) solve the system graphically. $$\left\\{\begin{array}{c}{x^{2}+y^{2}=25} \\ {3 x^{2}-16 y=0}\end{array}\right.$$

Step-by-Step Solution

Answer

The solutions are the points where the circle and parabola intersect. Exact coordinates would require calculation depending on the specific graphical representation or the use of an algebraic method.

Step 1: Graph the Circle Equation

The first equation describes a circle with a radius of \(5\) (since \(\sqrt{25}=5\)) and centered at the origin \((0,0)\). The graph of this equation can be drawn by marking the center and drawing a circle with a radius of \(5\) units.

Step 2: Graph the Parabola Equation

The second equation can be rearranged as \(y = 3x^{2}/16\). The graph of this equation describes a parabola with the vertex at the origin and opens upwards. Plot this graph on the same set of axes as the circle.

Step 3: Find Intersection Points

Intersection points are the points that are on both graphs simultaneously, and they represent the solutions to the system of equations. These are the points where the graphs of the circle and the parabola intersect. Visually identify or use algebraic methods to find these points. In general, if the system is complex, it might be more efficient to use algebraic methods to accurately find the intersection points.

Step 4: Confirm Solutions

Finally confirm the solutions by substituting the coordinates of intersection points into both original equations, ensuring that both equations are satisfied. This will confirm that found solutions are correct.

Key Concepts

Circle Equation

A circle equation is a mathematical way to describe all the points that are a fixed distance, known as the radius, from a central point called the center. In its standard form, the equation of a circle with a center at the origin \((h, k)\) and radius \(r\) is given by:

• \( (x - h)^2 + (y - k)^2 = r^2 \)

In the exercise example, the circle equation is \( x^2 + y^2 = 25 \). Here, the center is at the origin \((0,0)\) and the radius is 5, since \( \sqrt{25} = 5\).
This means every point \((x, y)\) on the circle is exactly 5 units away from the origin. To graph it, you start at the center point, \((0,0)\), and stretch out to a distance of 5 units in all directions, forming a circle.
Graphically, this creates a perfect round shape centered at the origin which is essential in solving graphical problems.

Parabola Equation

A parabola is a symmetrical open plane curve formed by points equidistant from a fixed point (focus) and a straight line (directrix). The standard form of a parabola equation is:

• \( y = ax^2 + bx + c \)

In this particular exercise, the parabola equation is derived from \(3x^2 - 16y = 0\), which we can rearrange to get \(y = \frac{3}{16}x^2\).
This equation indicates a parabola opening upwards, with its vertex at the origin \((0, 0)\). The factor \(\frac{3}{16}\) determines the "width" or "steepness" of the parabola. Graphing this involves plotting several \((x, y)\) points by choosing values for \(x\) and calculating \(y\), and then drawing a smooth curve through these points. The parabola will appear as a U-shaped curve centered vertically around the y-axis.

Intersection of Graphs

The intersection of graphs refers to the points where two different functions meet on a coordinate plane. In a system of equations like ours, finding these intersections graphically involves plotting both equations on the same set of axes.

• For the given exercise, it involves plotting the circle \(x^2 + y^2 = 25\) and the parabola \(y = \frac{3}{16}x^2\) on the same graph.

These intersection points represent solutions to the system, as they satisfy both equations simultaneously.
Finding these points visually can sometimes be imprecise, so confirming them by algebraically checking the coordinates against the original equations is crucial.Observing the graphs, these intersections occur at specific coordinates where both shapes overlap. Accurately identifying or calculating these points provides the solution to the system of equations. It's a process that combines both graphical and algebraic methods for verification.