Problem 91 - A Graphical Approach to Precalculus with Limits

Question

Shade the region(s) contained inside the graphs and give any points of intersection of the equations. $$y=2 x-1$$ $$y=2-x^{2}$$

Step-by-Step Solution

Answer

Points of intersection: (-3, -7) and (1, 1). Shade region between curves from x = -3 to x = 1.

Step 1: Set Equations Equal

We start by finding the points of intersection by equating the two equations: \(2x - 1 = 2 - x^2\).

Step 2: Rearrange and Simplify

Rearrange the equation to bring all terms to one side: \(x^2 + 2x - 3 = 0\).

Step 3: Solve the Quadratic Equation

Factor the quadratic equation: \((x+3)(x-1) = 0\). This gives the solutions \(x = -3\) and \(x = 1\).

Step 4: Find Corresponding y Values

Substitute \(x = -3\) and \(x = 1\) back into both original equations to find \(y\) values. \(y = 2(-3) - 1 = -7\) and \(y = 2(1) - 1 = 1\). Both points of intersection are \((-3, -7)\) and \((1, 1)\).

Step 5: Shading the Region

The graphs of the lines are a parabola opening downwards and a straight line. Between \(x = -3\) and \(x = 1\), the region enclosed is between these two graphs (above \(y = 2x - 1\) and below \(y = 2 - x^2\)). Shade this region.

Key Concepts

Quadratic Equation

In this exercise, we encounter a quadratic equation as part of solving for the points of intersection between two curves. The quadratic equation in question arises when we equate the given equations:

• The linear equation: \( y = 2x - 1 \)
• The quadratic equation: \( y = 2 - x^2 \)

To find where these curves intersect, set \( 2x - 1 = 2 - x^2 \).
This results in rearranging the terms to the form of a standard quadratic equation: \[ x^2 + 2x - 3 = 0 \]
Quadratic equations can be solved through factoring, as shown, but they can also be solved using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = 2 \), and \( c = -3 \). In this case, factoring is straightforward: \( (x+3)(x-1) = 0 \).
This provides the solutions \( x = -3 \) and \( x = 1 \), representing the x-coordinates of our points of intersection. Quadratic equations like these frequently appear in scenarios involving geometric shapes, making them a key concept in analyzing curves.

Graphical Solutions

In mathematics, graphical solutions involve using a graph to find intersections or other important points between functions. For this problem, we are interested in the intersection of the linear equation \( y = 2x - 1 \) and the quadratic equation \( y = 2 - x^2 \).
In practice, solving graphically involves plotting both equations on the same set of axes.

• The line \( y = 2x - 1 \) is straight and can be sketched by finding two points and drawing a line through them.
• The curve \( y = 2 - x^2 \) is a parabola opening downwards, recognizable by its negative coefficient in front of \( x^2 \).

By plotting these, you will see where they cross. The intersections occur at the solutions we found algebraically:

• \((-3, -7)\)
• \((1, 1)\)

These graphical solutions not only confirm our algebraic results but also provide a visual interpretation, helping us understand the spatial relationship between these two functions.

Shaded Regions

Shaded regions are areas on a graph contained between curves. In this problem, we're looking for the region between the line \( y = 2x - 1 \) and the parabola \( y = 2 - x^2 \).
First, identify the intersection points, which we've determined here to be \((-3, -7)\) and \((1, 1)\).

• For \( x \) values between these points, you need to determine which curve is "above" and which is "below" on the graph.
• In our case, the parabola \( y = 2 - x^2 \) is above the line \( y = 2x - 1 \) between \( x = -3 \) and \( x = 1 \).

To shade the region, simply fill in the area bounded above by the parabola and below by the line within these x-limits.
The shaded region helps in applications, such as determining areas under curves (integration in calculus) or visualizing solutions in constrained optimization problems.