Question

plot the graphs of both equations on the same coordinate plane. Find and label the points of intersection of the two graphs. $$ \begin{array}{l} y-3 x=1 \\ x^{2}+2 x+y^{2}=15 \end{array} $$

Short answer

Intersection points: \((0.849, 3.547)\) and \((-1.649, -3.947)\).

Step-by-step solution

Rearrange the Linear Equation

The first equation given is \( y - 3x = 1 \). We need to solve for \( y \) to make it easier to plot. Add \( 3x \) to both sides to get \( y = 3x + 1 \).

Identify the Quadratic Curve Equation

The second equation is \( x^2 + 2x + y^2 = 15 \), which is a quadratic equation involving both \( x \) and \( y \), indicating it is a circle or an ellipse depending on terms. This curve will be plotted as is.

Plot Linear Equation on Coordinate Plane

Now, plot the linear equation \( y = 3x + 1 \) on the coordinate plane. You can find points by choosing values for \( x \) and calculating corresponding \( y \) values. For example, when \( x = 0 \), \( y = 1 \).

Plot Quadratic Curve on Coordinate Plane

Plot the curve formed by \( x^2 + 2x + y^2 = 15 \). This can be done by recognizing it as a circle or ellipse and finding the center and radius or simply plotting multiple points that satisfy the equation.

Find Points of Intersection

To find where these graphs intersect, substitute \( y = 3x + 1 \) into the quadratic equation: \( x^2 + 2x + (3x + 1)^2 = 15 \). Simplify and solve for \( x \).

Solve the Intersection Equation

After substitution, simplify: \( x^2 + 2x + (9x^2 + 6x + 1) = 15 \), leading to \( 10x^2 + 8x + 1 = 15 \), which simplifies to \( 10x^2 + 8x - 14 = 0 \). Divide the whole equation by 2: \( 5x^2 + 4x - 7 = 0 \).

Solve Quadratic for x

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 5, b = 4, c = -7 \). Calculate the discriminant: \( b^2 - 4ac = 4^2 - 4 \cdot 5 \cdot (-7) = 16 + 140 = 156 \). Find possible \( x \) values.

Calculate x-values Using Quadratic Formula

Substitute into the quadratic formula: \( x = \frac{-4 \pm \sqrt{156}}{10} \). Calculate \( \sqrt{156} \approx 12.49 \) so \( x_1 \approx \frac{-4 + 12.49}{10} \approx 0.849 \) and \( x_2 \approx \frac{-4 - 12.49}{10} \approx -1.649 \).

Find Corresponding y-values

For each \( x \) value, substitute back into the linear equation \( y = 3x + 1 \). For \( x_1 \approx 0.849 \), \( y_1 = 3(0.849) + 1 \approx 3.547 \); for \( x_2 \approx -1.649 \), \( y_2 = 3(-1.649) + 1 \approx -3.947 \).

Plot and Label Points of Intersection

On the graph, plot both intersections points \( (0.849, 3.547) \) and \( (-1.649, -3.947) \). These are the points where the line intersects the curve.

Key concepts

Linear Equations

Linear equations are mathematical expressions that describe straight lines on a coordinate plane. They are typically written in the form \( y = mx + c \), where \( m \) is the slope of the line and \( c \) is the y-intercept, where the line crosses the y-axis. In our exercise, the linear equation is \( y = 3x + 1 \). This means the slope \( m \) is 3, indicating the line rises by 3 units for every 1 unit it moves to the right. The y-intercept is 1, so the line crosses the y-axis at the point \( (0, 1) \).

To plot this line on the coordinate plane, we start at the y-intercept, (0, 1), and use the slope to find another point. For instance, from \( (0, 1) \), you can move up 3 units and over 1 unit to the right to find the next point \( (1, 4) \). Connect these points to create the linear graph.

Quadratic Equations

Quadratic equations form curves called parabolas when plotted on a coordinate plane. These equations typically have the form \( ax^2 + bx + c = 0 \). The special thing about quadratic equations is that they show the relationship between \( x \) and \( y \) in a way that is more complex than a simple straight line.

In our exercise, we have the equation \( x^2 + 2x + y^2 = 15 \). This is not a simple quadratic equation involving just \( x \), but one that includes both \( x \) and \( y \), pointing towards circular or elliptical paths depending on its terms. In this case, it represents a shape similar to a circle or an ellipse because both \( x \) and \( y \) are squared. To plot this shape, multiple solutions in terms of pairs of \((x, y)\) are plotted that satisfy the equation, forming a closed curve.

Coordinate Plane

A coordinate plane is a two-dimensional surface on which points, lines, and curves can be plotted. It is defined by a horizontal axis, known as the x-axis, and a vertical axis, known as the y-axis. The intersection of these axes is called the origin, represented as \((0, 0)\). Each point on this plane has coordinates \((x, y)\), showing its position relative to the origin.

In this exercise, both the linear equation and the quadratic curve are plotted on the coordinate plane. Doing so allows us to visualize the intersections of these two different types of graphs. The linear graph is a straight line, while the quadratic curve forms a more rounded shape. Where these two meet is crucial for solving real-world problems, as those points represent the solutions that satisfy both equations in the system.

Points of Intersection

The points of intersection are where two graphs meet or cross each other on the coordinate plane. In other words, these are the pairs of \((x, y)\) that satisfy both equations simultaneously. Finding these points is crucial for solving systems of equations, as it provides the answer to when and where the conditions described by both equations occur together.

In our exercise, we have found that the line described by \( y = 3x + 1 \) intersects with the curve described by \( x^2 + 2x + y^2 = 15 \) at two points. These are approximately at \((0.849, 3.547)\) and \((-1.649, -3.947)\). To find these points, we substituted the linear equation into the quadratic to reduce the system to a single equation with one variable, then solved for \( x \). Once \( x \) values were found, they were substituted back into the linear equation to find the corresponding \( y \), giving us the exact points of intersection.