Problem 94 - 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. $$\begin{aligned}&y=2 x^{2}+x-3\\\&y=x^{2}-2 x+1\end{aligned}$$

Step-by-Step Solution

Answer

Intersection points are (-4, 21) and (1, 0). Shade between the curves within these x-values.

Step 1: Equate the Equations

To find the points of intersection, set the equations equal to each other: \(2x^2 + x - 3 = x^2 - 2x + 1\). Simplify this to find \(x\).

Step 2: Rearrange the Equations

Rearrange the equation obtained in Step 1: \(2x^2 + x - 3 - x^2 + 2x - 1 = 0\). Simplifying gives \(x^2 + 3x - 4 = 0\).

Step 3: Factor the Quadratic

Factor the quadratic equation \(x^2 + 3x - 4 = 0\). This can be rewritten as \((x + 4)(x - 1) = 0\).

Step 4: Solve for x

From the factors obtained, solve \(x + 4 = 0\) and \(x - 1 = 0\). This gives the solutions \(x = -4\) and \(x = 1\).

Step 5: Find Corresponding y-values

Substitute \(x = -4\) and \(x = 1\) back into either original equation to find \(y\). Using \(y = x^2 - 2x + 1\), for \(x = -4\), \(y = 21\); and for \(x = 1\), \(y = 0\).

Step 6: Points of Intersection

The points of intersection are \((-4, 21)\) and \((1, 0)\).

Step 7: Analyze the Graphs

Graph both parabolas. The first equation, \(y = 2x^2 + x - 3\), opens upwards with a vertex below the other parabola \(y = x^2 - 2x + 1\) which also opens upwards. The regions inside the intersection points are between these graphs.

Step 8: Shading the Region

Shade the region between the curves, focusing on the area bounded between the points \((x = -4)\) and \((x = 1)\). The relevant region lies between the graphs of the two parabolas between these x-values.

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation that are very useful in many areas of mathematics and applied science. They are generally represented in the standard form: \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants, and \( a \) is not zero.These equations are called "quadratic" because "quad" refers to the number four, relating to the degree or power of two in the equation (since a degree of two means the equation is quadratic, not linear or cubic). Quadratic equations can be solved by various methods:

• Factoring
• Completing the Square
• Using the Quadratic Formula
• Graphing

In this exercise, we used factoring to solve the quadratic equation formed by equating two parabolas. This helps us determine the x-values where the graphs intersect, which are crucial for understanding the relationship between the two equations.

Points of Intersection

The points of intersection are where two or more graphs meet or cross each other. Finding these points is a matter of solving the equations simultaneously, which effectively means finding common solutions that satisfy both equations. In this context, when two graphs of parabolas intersect, we set their respective equations equal to determine the x-values of these points. The reason this works is that, at the intersection points, both equations give the same y-value for a particular x-value. For example, by setting the given quadratic equations \( 2x^2 + x - 3 \) and \( x^2 - 2x + 1 \) equal to each other, we derive a new quadratic equation: \( x^2 + 3x - 4 = 0 \). After solving, we find the x-values of intersection as \( x = -4 \) and \( x = 1 \). Once we have the x-values, substituting back into either original equation gives the respective y-values, resulting in the points of intersection: \((-4, 21)\) and \((1, 0)\). It's essential to get these points right as they help in drawing the graph accurately and shading the correct region.

Parabolas

Parabolas are the U-shaped graphs that represent quadratic equations. Each quadratic equation can be transformed into a parabola on a Cartesian coordinate plane. The form of the parabola is determined by the coefficient of \( x^2 \). If this coefficient is positive, the parabola opens upwards; if negative, it opens downwards.For instance, the equation \( y = 2x^2 + x - 3 \) is a parabola that opens upwards. The vertex of this parabola is at a point beneath the vertex of the other parabola \( y = x^2 - 2x + 1 \), which also opens upwards.When dealing with parabola graphs in exercises like this one, it's beneficial to comprehend:

• Vertex: The highest or lowest point of a parabola.
• Axis of symmetry: The vertical line that divides the parabola into two mirror-image halves.
• Direction: Whether the parabola opens upwards or downwards affects the area between it and another parabola.

In our case, both parabolas are intersected between the x-values \(x = -4\) and \(x = 1\). Understanding the direction and shape helps in shading the area correctly, which represents the region enclosed between these parabolas.