Problem 3 - A Graphical Approach to Precalculus with Limits

Question

In Exercises begin by drawing a rough sketch to determine the number of real solutions for the equation \(y_{1}=y_{2}\) Then solve this equation by hand. Give the solution set and any extraneous values that may occur. Do not use a calculator. $$\begin{aligned} &y_{1}=\sqrt{x}\\\ &y_{2}=-x+3 \end{aligned}$$

Step-by-Step Solution

Answer

The equation has two real solutions: \(x = \frac{7 - \sqrt{13}}{2}\) and \(x = \frac{7 + \sqrt{13}}{2}\).

Step 1: Sketch the Graph

First, let's sketch the graphs of the equations. The graph of \(y_1 = \sqrt{x}\) is the right-half of a parabola opening upwards, starting at the origin (0,0) and extending to the right. The graph of \(y_2 = -x + 3\) is a straight line with a negative slope intersecting the y-axis at the point (0,3). By plotting these graphs, you can visually estimate where they might intersect.

Step 2: Set Equations Equal to Each Other

Set \(y_1\) equal to \(y_2\) to find the points of intersection. This gives the equation: \[ \sqrt{x} = -x + 3 \]

Step 3: Solve the Equation

Square both sides to eliminate the square root:\[ (\sqrt{x})^2 = (-x + 3)^2 \]Simplify:\[ x = x^2 - 6x + 9 \]

Step 4: Rearrange and Factor the Quadratic

Rearrange the equation to form a standard quadratic:\[ x^2 - 7x + 9 = 0 \] Now, factor this quadratic equation. Try to find two numbers that multiply to 9 and add up to -7. This can be solved using the quadratic formula if factoring is difficult.

Step 5: Apply the Quadratic Formula

Since factoring isn't straightforward, apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]With \(a = 1\), \(b = -7\) and \(c = 9\), plug in these values:\[ x = \frac{7 \pm \sqrt{49 - 36}}{2} \]\[ x = \frac{7 \pm \sqrt{13}}{2} \]

Step 6: Evaluate the Roots

The solutions, \( x = \frac{7 + \sqrt{13}}{2} \) and \( x = \frac{7 - \sqrt{13}}{2} \), provide potential points of intersection. We must check for extraneous solutions.

Step 7: Check the Validity of Solutions

Since the original equation involved a square root, verify the solutions in the context of \(y_1 = \sqrt{x}\), which is only defined for \(x \geq 0\). Calculate the approximate values: \( \frac{7 - \sqrt{13}}{2} \approx 1.697 \), and \( \frac{7 + \sqrt{13}}{2} \approx 5.303 \). Only positive values are valid for \(y_1\), so both are valid.

Step 8: Conclusion on Real Solutions

Both solutions \( \frac{7 - \sqrt{13}}{2} \) and \( \frac{7 + \sqrt{13}}{2} \) are valid since they satisfy the requirement of \(x\) being non-negative. Therefore, both of these points are valid points of intersection between the curves.

Key Concepts

Graphical Analysis

When dealing with functions, graphical analysis is a useful tool to predict and identify possible solutions. By plotting a graph, you can visually assess the relationship between different equations, like comparing where they might intersect. For the equations \(y_1 = \sqrt{x}\) and \(y_2 = -x + 3\), sketching helps in visualizing the behavior of these equations.
For example:

• The function \(y_1 = \sqrt{x}\) represents a graph that starts at the origin and curves upwards to the right. It's essentially the upper part of a sideways parabola.
• The function \(y_2 = -x + 3\) results in a straight line with a negative slope. The line intercepts the vertical (y) axis at 3, sloping downward.

From the graph, you can make educated guesses about where these two graphs might intersect without solving the equations simultaneously at the first step.

Quadratic Equation

The quadratic equation is central to solving many problems in mathematics, due to its structure: \(ax^2 + bx + c = 0\). This form can be manipulated to find solutions or solve for unknown values. In our problem, once the equations are set equal, you get a quadratic equation.

• Initially, by equating \(y_1\) and \(y_2\), you form the equation \(\sqrt{x} = -x + 3\).
• By squaring both sides, you transform it into a quadratic form, \(x = x^2 - 6x + 9\).
• Reorganizing it gives \(x^2 - 7x + 9 = 0\), a standard quadratic equation.

It's important because not all equations are easily factorable, and sometimes we'll need to use methods like the quadratic formula to find our solutions.

Square Root Function

The square root function, denoted as \(\sqrt{x}\), characterizes one of the two functions we are analyzing. It is important to understand its properties when predicting the behavior of your solutions.

• The function \(\sqrt{x}\) is only defined for non-negative values of \(x\), meaning \(x \geq 0\).
• Graphically, it appears as a curve starting at the origin (0,0) and moves upwards and to the right.

Because of its definition, any solutions to a problem involving \(\sqrt{x}\) need to meet this non-negativity constraint for \(x\). This requirement becomes key when validating the roots calculated from our quadratic equation.

Intersection of Graphs

The intersection of graphs is a critical concept as it provides the solution to equations where two functions meet. In mathematical terms, it identifies the common solutions to two separate equations denoted as \(y_1 = y_2\).

• To find these points, set the expressions for \(y_1\) and \(y_2\) equal, thus forming an equation that can be solved further to find points of intersection.
• After solving \(\sqrt{x} = -x + 3\), we derived \(x = \frac{7 \pm \sqrt{13}}{2}\), which were approximations of potential intersection points.
• These solutions must also adhere to the conditions of function definitions, like the non-negativity for \(y_1 = \sqrt{x}\).

Confirming both solutions are valid by recalculating them under the square root function's requirement (\(x\geq0\)) ensures they represent true intersections.