Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 31-38, plot the graphs of both equations on the same coordinate plane. Find and label the points of intersection of the two graphs (see Example 4). $$ \begin{aligned} &y-3 x=1 \\ &x^{2}+2 x+y^{2}=15 \end{aligned} $$

Step-by-Step Solution

Answer

The intersection points are \( \left( \frac{-2 + \sqrt{39}}{5}, \frac{-1 + 3\sqrt{39}}{5} \right) \) and \( \left( \frac{-2 - \sqrt{39}}{5}, \frac{-1 - 3\sqrt{39}}{5} \right) \).

Step 1: Rearrange the first equation

The first equation is given as \( y - 3x = 1 \). We need to express \( y \) in terms of \( x \):\[ y = 3x + 1 \].

Step 2: Identify the second equation

The second equation is \( x^2 + 2x + y^2 = 15 \). This is the equation of a circle in standard form, \((x + 1)^2 + y^2 = 16\). The circle has a center at (-1, 0) and a radius of 4.

Step 3: Substitute the rearranged equation into the second equation

Substitute \( y = 3x + 1 \) into \( (x + 1)^2 + y^2 = 16 \) to find the intersection points:\[(x + 1)^2 + (3x + 1)^2 = 16\].

Step 4: Simplify and solve for x

Expand and simplify the equation:\[(x + 1)^2 + (3x + 1)^2 = 16\]\[x^2 + 2x + 1 + 9x^2 + 6x + 1 = 16\]\[10x^2 + 8x + 2 = 16\]Move all terms to one side and simplify:\[10x^2 + 8x + 2 - 16 = 0\]\[10x^2 + 8x - 14 = 0\].

Step 5: Factor the quadratic or use the quadratic formula

The quadratic equation obtained is \( 10x^2 + 8x - 14 = 0 \). Simplify before using the quadratic formula:\[5x^2 + 4x - 7 = 0\].Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 5 \), \( b = 4 \), \( c = -7 \):\[x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 5 \cdot (-7)}}{10}\]\[x = \frac{-4 \pm \sqrt{16 + 140}}{10}\]\[x = \frac{-4 \pm \sqrt{156}}{10}\].

Step 6: Calculate the values for x

Simplifying further, we find:\[x = \frac{-4 \pm 2\sqrt{39}}{10}\]\[x = \frac{-2 \pm \sqrt{39}}{5}\].

Step 7: Find corresponding y values

Substitute the \( x \) values back into \( y = 3x + 1 \):For \( x = \frac{-2 + \sqrt{39}}{5} \):\[ y = 3 \left( \frac{-2 + \sqrt{39}}{5} \right) + 1 \]\[ y = \frac{-6 + 3\sqrt{39}}{5} + 1 \]\[ y = \frac{-1 + 3\sqrt{39}}{5} \].For \( x = \frac{-2 - \sqrt{39}}{5} \):\[ y = 3 \left( \frac{-2 - \sqrt{39}}{5} \right) + 1 \]\[ y = \frac{-6 - 3\sqrt{39}}{5} + 1 \]\[ y = \frac{-1 - 3\sqrt{39}}{5} \].

Step 8: State the points of intersection

The points of intersection are:\[ \left( \frac{-2 + \sqrt{39}}{5}, \frac{-1 + 3\sqrt{39}}{5} \right) \]and \[ \left( \frac{-2 - \sqrt{39}}{5}, \frac{-1 - 3\sqrt{39}}{5} \right) \].

Key Concepts

Intersection Points

When we talk about intersection points in systems of equations, we mean the locations on the coordinate plane where the graphs of the equations meet. For two equations, these are the unique solutions that satisfy both equations simultaneously. In our problem, the intersection points are where the line, defined by the equation \( y = 3x + 1 \), crosses the circle \((x + 1)^2 + y^2 = 16\).

Finding these points typically involves solving a system of equations. We substitute and manipulate the equations to reveal the exact coordinates of these points. In this case, after solving, the intersection points were found to be \( \left( \frac{-2 + \sqrt{39}}{5}, \frac{-1 + 3\sqrt{39}}{5} \right) \) and \( \left( \frac{-2 - \sqrt{39}}{5}, \frac{-1 - 3\sqrt{39}}{5} \right) \).

Understanding how intersection points work helps in realizing the relationship between different functions and shapes on a plane.

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of any quadratic equation, which is a polynomial of the form \( ax^2 + bx + c = 0 \). This formula gives you the values of \( x \) that satisfy the equation. It is expressed as:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In the exercise, the quadratic equation \( 5x^2 + 4x - 7 = 0 \) was solved using the quadratic formula.

The steps to use the formula involve:

• Identifying the coefficients \( a \), \( b \), and \( c \).
• Substituting these into the formula.
• Calculating the discriminant (the part under the square root).
• Finding the two potential solutions.

In our example, this method allowed us to precisely determine the \( x \)-values at the points of intersection.

Graphing Equations

Graphing equations is vital in visualizing how different mathematical relationships appear in a coordinate plane. It helps us identify intersection points, compare slopes, and analyze trends or behavior of functions. In our exercise, graphing both the line \( y = 3x + 1 \) and the circle \((x + 1)^2 + y^2 = 16\) shows their points of intersection.

Essential steps when graphing:

• Identify the type of graph (line, circle, parabola, etc.)
• Find key points (like intercepts or center for a circle).
• Plot multiple points to ensure accuracy.
• Connect these points smoothly for the curve or line.

Graphing not only visualizes the solution but enhances our understanding of the relationships governing the equations.

Coordinate Geometry

Coordinate geometry, or analytic geometry, deals with placing geometric figures in a numerical framework on the coordinate plane. It's a blend of algebra and geometry, and in this context, it allows us to describe geometric shapes with equations and find solutions analytically.

In our task, it was used to locate the exact meeting points of a line and a circle by solving equations in the coordinate plane. By setting the line's equation equal to the circle's expression through substitution and simplification, we derived specific intersection points.

This powerful method not only provides precise solutions but also connects various forms of mathematical representation—be they geometric, algebraic, or numerical—making it invaluable for problem-solving in math.