Problem 15 - Algebra 2 and Trigonometry

Question

In \(12-17,\) use a graph to find the solution set of each inequality. $$ x^{2}-2 x+1 < 0 $$

Step-by-Step Solution

Answer

The solution set is an empty set; there are no real solutions.

Step 1: Identify Type of Inequality

The inequality given is \(x^{2}-2x+1 < 0\). Notice that this is a quadratic inequality, which can be graphically represented by a parabola. The expression \(x^{2}-2x+1\) is equivalent to \((x-1)^{2}\), which is a perfect square.

Step 2: Determine Critical Points

To determine where \((x-1)^2 < 0\), we first identify where the expression equals zero, i.e., find the roots. Solving the equation \((x-1)^{2} = 0\) gives \(x = 1\). This is our critical point where the graph touches the x-axis.

Step 3: Analyze the Sign Changes

Since \((x-1)^2\) is a perfect square, the parabola resulting from this equation is always non-negative and only zero at \(x = 1\). Therefore, the expression \((x-1)^{2} < 0\) has no real solutions because squares are always non-negative and only zero at the point we identified.

Step 4: Graphical Interpretation

Graph the function \(y = (x-1)^2\). It's a parabola opening upwards with its vertex at \((1,0)\). Since the parabola is non-negative for all \(x\)-values and equals zero only at the vertex, it never goes below the x-axis.

Step 5: Conclusion about the Inequality

Since \((x-1)^2\) is never less than zero, the solution set where \((x-1)^{2} < 0\) is the empty set, which means there are no \(x\)-values that satisfy this inequality.

Key Concepts

Parabola

A parabola is a U-shaped curve that represents a quadratic function like the one given in the inequality problem: \(x^2 - 2x + 1 < 0\). When graphed, a parabola can open upwards or downwards depending on the coefficient of the \(x^2\) term.
In our exercise, the expression \((x-1)^2\) represents a parabola that opens upwards. This is because the coefficient in front of the \(x^2\) term is positive. The shape of this parabola is symmetric around its vertex, a crucial property when analyzing solutions of inequalities.

• A positive coefficient means the parabola opens upwards, forming a smiley shape.
• A negative coefficient means the parabola opens downwards, forming a frowny shape.
• The vertex is the highest or lowest point on the parabola, depending on its orientation.

Understanding these basic properties of parabolas is key to graphically representing quadratic inequalities.

Critical Points

Critical points are points on the graph where important changes occur. In quadratic inequalities, critical points are typically where the expression equals zero. For \((x-1)^2\), solving \((x-1)^2 = 0\) yields the critical point at \(x = 1\). This is the vertex of the parabola.
The critical point of a parabola in terms of inequalities signifies where the curve touches or crosses the x-axis. Since \((x-1)^2\) is a perfect square and always non-negative, the critical point \(x = 1\) is where the parabola touches the x-axis without crossing it.

• Identify critical points by setting the quadratic expression to zero.
• Critical points often serve as x-intercepts for the graph.
• They provide valuable information about the inequality's solution set.

Graphical Interpretation

Graphically interpreting quadratic inequalities involves sketching the quadratic function and analyzing the graph concerning the x-axis.
For the inequality \((x-1)^2 < 0\), we graph \(y = (x-1)^2\). This parabola has a vertex at \((1,0)\) and opens upwards.

• This parabola never dips below the x-axis, as it's non-negative for all \(x\).
• The inequality \((x-1)^2 < 0\) suggests that we look for \(x\)-values where the curve is below the x-axis.
• Since it never goes below, there are no \(x\)-values that satisfy this inequality, resulting in an empty set.

Visualizing the graph clarifies that the parabola's position relative to the x-axis directly answers the inequality. This is why developing skills in sketching and interpreting graphs is essential for solving quadratic inequalities.