Problem 39 - Precalculus: Mathematical for Calculus

Question

Use the graphical method to find all solutions of the system of equations, correct to two decimal places. $$\left\\{\begin{array}{l}y=x^{2}+8 x \\\y=2 x+16\end{array}\right.$$

Step-by-Step Solution

Answer

The solutions are \((2, 20)\) and \((-8, 0)\).

Step 1: Identify the Equations

We are given two equations in the system: 1. \( y = x^2 + 8x \) 2. \( y = 2x + 16 \)The goal is to find the points where these two equations intersect graphically, which are the solutions to the system.

Step 2: Graph the Quadratic Equation

Graph the quadratic equation \( y = x^2 + 8x \). This is a parabola opening upwards with the vertex form given by completing the square or using the vertex formula. Break down the parabola's vertex by finding \( x = - \frac{b}{2a} = - \frac{8}{2} = -4 \). Substitute back to find the vertex point (-4, -16).

Step 3: Graph the Linear Equation

Graph the linear equation \( y = 2x + 16 \). This is a straight line with a slope of 2 and a y-intercept at 16. Plot the y-intercept (0, 16) and use the slope to find another point, such as (1, 18).

Step 4: Determine Intersection Points

On the graph, observe where the parabola and the line intersect. These points of intersection can be considered the solutions. These intersections occur when both equations have the same \( y \) value for the same \( x \) value.

Step 5: Solve Algebraically for Exact Intersection

Set the equations equal to each other to find the exact points:\[ x^2 + 8x = 2x + 16\]Rearrange to form a quadratic equation:\[ x^2 + 6x - 16 = 0\]Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) Here, \( a=1, b=6, c=-16 \).

Step 6: Solve with Quadratic Formula

Substitute into the quadratic formula:\[ x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-16)}}{2(1)}\]\[ x = \frac{-6 \pm \sqrt{36 + 64}}{2}\]\[ x = \frac{-6 \pm \sqrt{100}}{2}\]\[ x = \frac{-6 \pm 10}{2}\]This gives \( x = 2 \) or \( x = -8 \).

Step 7: Find Corresponding y-Values

Substitute \( x = 2 \) and \( x = -8 \) back into either original equation (e.g., \( y = 2x + 16 \)) to find corresponding \( y \)-values:For \( x = 2 \), \( y = 2(2) + 16 = 20 \).For \( x = -8 \), \( y = 2(-8) + 16 = 0 \).The solutions are \((2, 20)\) and \((-8, 0)\).

Key Concepts

System of Equations

A *system of equations* consists of two or more equations with the same set of unknowns. When solving a system, we're looking for values of the variables that satisfy all the equations simultaneously. In this case, our system involves two equations involving variables "x" and "y":

• The quadratic equation: \( y = x^2 + 8x \)
• The linear equation: \( y = 2x + 16 \)

The primary goal is to identify the points at which both equations are true at the same time. Graphically, these solutions are represented by the intersection points of their respective graphs. This method gives a clear visual representation of where the lines or curves meet.

Quadratic Equation

A *quadratic equation* is a polynomial equation in which the highest degree term is of second degree. It is often written in the standard form: \( ax^2 + bx + c = 0 \). Here, it takes the form \( y = x^2 + 8x \), where:

• \( a = 1 \)
• \( b = 8 \)
• \( c = 0 \)

This equation produces a parabola when graphed. The shape of the graph depends on the values of "a", "b", and "c". In our equation, the parabola opens upwards because "a" is positive.

To graph this equation, a helpful method is to find the vertex of the parabola, which is the lowest point in this case. This can be found using the formula \( x = -\frac{b}{2a} \). For our equation, the vertex is at the point \( (-4, -16) \). Graphing involves calculating a few points on either side of the vertex and connecting them with a smooth curve.

Intersection Points

The *intersection points* are where the graphs of two or more equations meet. These points signify solutions to the system of equations, as they satisfy all equations involved simultaneously.

To find these graphically, plot both equations on the same set of axes. The linear equation \( y = 2x + 16 \) is a straight line, while the quadratic equation \( y = x^2 + 8x \) forms a parabola. The intersection points are where these two graphs overlap.

• For the given system, upon setting the equations equal to each other, we solve: \( x^2 + 8x = 2x + 16 \).
• Simplifying this leads to the quadratic \( x^2 + 6x - 16 = 0 \).

Using the quadratic formula gives us solutions \( x = 2 \) and \( x = -8 \). By substituting back, we confirm the corresponding \( y \)-values, resulting in intersection points \((2, 20)\) and \((-8, 0)\). These are the solutions to the system of equations.