Problem 92 - 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. $$y=x^{2}-x+1$$ $$y=-x^{2}+1$$

Step-by-Step Solution

Answer

Intersection points are (0, 1) and (0.5, 0.75); region from x=0 to x=0.5 is shaded.

Step 1: Set the Equations Equal

To find the points of intersection, set the two equations equal to each other: \[-x^2 + 1 = x^2 - x + 1\].By solving this equation, we can determine where the graphs intersect.

Step 2: Simplify the Equation

Combine like terms to simplify the equation: \[0 = 2x^2 - x\].This simplifies to: \[2x^2 - x = 0\].

Step 3: Factor the Quadratic Equation

Factor the simplified equation to find the possible values of \(x\): \[x(2x - 1) = 0\].Setting each factor equal to zero gives us \(x = 0\) and \(2x - 1 = 0\).

Step 4: Solve for x

Solve the factored equation for \(x\):\(x = 0\) is already solved.For \(2x - 1 = 0\), solve for \(x\) to get \(x = \frac{1}{2}\).

Step 5: Find Corresponding y Values

Substitute \(x = 0\) and \(x = \frac{1}{2}\) into either original equation to find the corresponding \(y\) values.For \(x = 0\):\[y = -0^2 + 1 = 1\].For \(x = \frac{1}{2}\):\[y = -\left(\frac{1}{2}\right)^2 + 1 = -\frac{1}{4} + 1 = \frac{3}{4}\].

Step 6: Determine Regions for Shading

To find the regions to shade, analyze the inequality created by comparing the two equations.For the ranges of \(x\) between the points of intersection (0 and \(\frac{1}{2}\)), determine which graph is greater: \For values of \(x\) less than 0 and greater than \(\frac{1}{2}\), analyze the sign of the inequality (-\(x^2+1) > (x^2-x+1)\).

Step 7: Verify by Testing a Point

Test a point between \(x = 0\) and \(x = \frac{1}{2}\):Use \(x = 0.25\):Calculate using both equations to determine which one is greater.Check if (-\(0.25^2 + 1 \approx 0.9375\)) is greater than (\(0.25^2 - 0.25 + 1 \approx 0.9375\))

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in mathematics. They often appear in the form:

• \(y = ax^2 + bx + c\)

where \(a\), \(b\), and \(c\) are constants and \(x\) is the variable. The distinctive feature of a quadratic equation is its highest degree term, \(x^2\) which gives its graph a parabola shape.
In this exercise, we have two different quadratic equations:

• \(y = x^2 - x + 1\)
• \(y = -x^2 + 1\)

Each equation represents a different parabola. The first is upwards opening due to the positive coefficient of \(x^2\), while the second opens downwards due to the negative \(x^2\) term. Understanding their forms allows us to solve effectively for intersections.

Graphical Representation

Graphical representation helps us visualize the solutions of equations. Every equation can be shown as a graph on a coordinate plane.
In the case of quadratic equations, these graphs are typically parabolas. The points where the graphs intersect represent the solutions to the equations when set equal, indicating values of \(x\) and \(y\) that satisfy both equations.
Plotting the equations

• \(y = x^2 - x + 1\)
• \(y = -x^2 + 1\)

shows two parabolas: one opening upwards and the other downwards. The graphical intersection points occur at specific \(x\)-coordinates, where their \(y\)-values match. These are the solutions to the equation when we set the equations equal.

Inequalities

Inequalities compare values to determine which is greater, less, or equal over specific ranges of \(x\). In this exercise, shading regions involves analyzing inequalities based on where one parabola is above or below the other.
The key inequality here is:

• \(-x^2 + 1 > x^2 - x + 1\)

This inequality implies that for some \(x\)-values, one equation is greater than the other. It involves exploring the "areas" between the two parabolas.
By testing values of \(x\) from the solution points, we identify the ranges where each function is greater and shade those regions. This creates a shaded region between the curves, marking the solutions to the inequalities.

Solution Verification

Solution verification ensures the answers are correct and applicable. Once we identify the points of intersection, verifying involves substituting back values to check their validity in the original equations.
For example, substituting \(x = 0\) and \(x = \frac{1}{2}\) into the original equations provide their corresponding \(y\)-values, ensuring these do indeed lie on both curves.
Moreover, the inequality verification step checks the selected shaded regions. Testing points like \(x = 0.25\) between the main intersections ensures the chosen region is accurately shaded. For instance, both calculations yield approximately 0.9375, confirming equal results between these \(x\)-range checks.
Through these methods, we can confidently conclude the regions and points marked indeed solve both the original and inequality equations.