Problem 56 - Algebra 2 and Trigonometry

Question

a. On the same set of axes, sketch the graphs of \(y=x^{2}-4 x+5\) and \(y=2 x+2\) b. Does the system of equations \(y=x^{2}-4 x+5\) and \(y=2 x+2\) have a common solution in the set of real numbers? Justify your answer. c. Find the solution set.

Step-by-Step Solution

Answer

The system has two real solutions: \((3+\sqrt{6}, 8+2\sqrt{6})\) and \((3-\sqrt{6}, 8-2\sqrt{6})\).

Step 1: Understand the Problem

Identify the key components of the equations. We have a quadratic equation: \(y = x^2 - 4x + 5\), which is a parabola opening upwards. The other equation is linear: \(y = 2x + 2\), which is a straight line.

Step 2: Graph the Quadratic Equation

The quadratic equation \(y = x^2 - 4x + 5\) represents a parabola. Find the vertex using the formula \(x = \frac{-b}{2a}\). Here, \(a = 1\) and \(b = -4\), so \(x = \frac{4}{2} = 2\). Substitute \(x = 2\) into the equation to find \(y\): \(y = 2^2 - 4(2) + 5 = 1\). The vertex is \((2, 1)\). The parabola passes through \((0, 5)\) as it's the y-intercept.

Step 3: Graph the Linear Equation

For the linear equation \(y = 2x + 2\), identify the y-intercept and the slope. The y-intercept is \((0, 2)\), and the slope is 2, meaning for every increase of 1 in \(x\), \(y\) increases by 2. Draw the line passing through \((0, 2)\) and using the slope to find another point, such as \((1, 4)\).

Step 4: Find the Intersection Points

Set the quadratic and linear equations equal to find their intersection points: \(x^2 - 4x + 5 = 2x + 2\). Rearrange to form the equation \(x^2 - 6x + 3 = 0\).

Step 5: Solve the Quadratic Equation

Solve \(x^2 - 6x + 3 = 0\) using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = -6\), and \(c = 3\). Calculate the discriminant: \((-6)^2 - 4(1)(3) = 36 - 12 = 24\). Since the discriminant is positive, there are two real solutions.

Step 6: Calculate the Roots

Compute \(x = \frac{6 \pm \sqrt{24}}{2}\). Simplifying gives \(x = \frac{6 \pm 2\sqrt{6}}{2}\), resulting in \(x = 3 \pm \sqrt{6}\). These give two intersection points.

Step 7: Find y for Each x

Calculate the y-values for the intersection points by substituting back into \(y = 2x + 2\). For \(x = 3 + \sqrt{6}\): \(y = 2(3 + \sqrt{6}) + 2 = 8 + 2\sqrt{6}\). For \(x = 3 - \sqrt{6}\): \(y = 2(3 - \sqrt{6}) + 2 = 8 - 2\sqrt{6}\). Thus, the points are \((3 + \sqrt{6}, 8 + 2\sqrt{6})\) and \((3 - \sqrt{6}, 8 - 2\sqrt{6})\).

Key Concepts

Solving Quadratic Equations

To solve a quadratic equation like \(x^2 - 4x + 5 = 0\), we need to find the values of \(x\) that make the equation true. Quadratic equations, typically of the form \(ax^2 + bx + c = 0\), can have up to two solutions because they represent parabolas which may cross the x-axis in up to two places.
To solve these equations, you have a few techniques at your disposal:

• Factoring: This involves expressing the quadratic in a product of two binomials. For example, \(x^2 - 4x + 4\) factors into \((x - 2)(x - 2)\). But not all quadratics are easily factorable, which leads us to the other methods.
• Completing the square: This method involves making the quadratic part of the expression a perfect square trinomial. This is slightly more complex and isn’t always the quickest method.
• Quadratic formula: This is a universal method that can solve any quadratic equation of the form \(ax^2 + bx + c = 0\), using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Using these methods helps us quickly and reliably determine where a quadratic function meets the x-axis if at all.

Intersection of Graphs

An intersection of graphs is a point where two or more graphs meet, representing values that are solutions to multiple equations. In the case of quadratic and linear equations, you often find two types of intersections: either none, one, or two.
When a parabola (quadratic) and a line (linear) intersect:

• No intersection: This occurs when the parabola and line are too far apart, so they never touch—this happens if the line is entirely above or below the parabola.
• One intersection: Here, the line is tangent to the parabola, touching it at exactly one point.
• Two intersections: This is the most common scenario where the line crosses the parabola at two distinct points.

To find these points for our equations \(y = x^2 - 4x + 5\) and \(y = 2x + 2\), we set them equal to each other: \(x^2 - 4x + 5 = 2x + 2\). Rearranging gives \(x^2 - 6x + 3 = 0\). Solving this quadratic equation enables us to find the x-values where intersections occur. These x-values can then be substituted back into either original equation to find the corresponding y-values, giving the intersection points.

Parabolas and Lines

Understanding the relationship between parabolas and lines is crucial for graphing and solving systems involving quadratic and linear equations. A parabola is a U-shaped curve that can open upwards or downwards, described by quadratic equations such as \(y = x^2 - 4x + 5\). This equation reveals a parabola opening upwards, with a vertex, which is its highest or lowest point.The line, represented by a linear equation such as \(y = 2x + 2\), is straight and described through its slope and y-intercept.

• Vertex of the parabola: The vertex can be found using the formula for the x-coordinate \(x = \frac{-b}{2a}\) and substituting it back into the original quadratic to find the y-coordinate.
• Slope of the line: This tells us how steep the line is. A slope of 2, as in our line \(y = 2x + 2\), means that for every unit increase in \(x\), \(y\) increases by 2.
• Y-intercept of the line: This is where the line crosses the y-axis, at point \(y = 2\) when \(x = 0\).

By graphing both, we can visually locate the intersection points or calculate them algebraically.

Quadratic Formula

The quadratic formula is a powerful tool used to find the solutions for any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula itself is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(b^2 - 4ac\) is known as the discriminant.
Understanding how the quadratic formula works:

• Discriminant: The expression \(b^2 - 4ac\) under the square root sign is crucial as it determines the number and type of solutions:
  • A positive discriminant indicates two distinct real solutions, meaning the parabola intersects the x-axis at two points.
  • A zero discriminant results in exactly one real solution, where the parabola just touches the x-axis.
  • A negative discriminant means no real solutions, indicating the quadratic does not intersect the x-axis at all.
• Calculation: The formula calculates the x-values at which these intersections occur, assuming real solutions exist. For our equation \(x^2 - 6x + 3 = 0\), the discriminant was \(24\) (positive), leading to two real solutions: \(x = 3 \pm \sqrt{6}\).

This method, being independent of the particular form or factorability of the quadratic, provides a consistent and reliable means to determine intersection points with the x-axis or to solve systems involving lateral mathematic elements.