Problem 54 - Precalculus: Mathematical for Calculus

Question

Intersection of a Parabola and a Line On a sheet of graph paper, or using a graphing calculator, draw the parabola \(y=x^{2} .\) Then draw the graphs of the linear equation \(y=x+k\) on the same coordinate plane for various values of \(k .\) Try to choose values of \(k\) so that the line and the parabola intersect at two points for some of your \(k\) 's, and not for others. For what value of \(k\) is there exactly one intersection point? Use the results of your experiment to make a conjecture about the values of \(k\) for which the following system has two solutions, one solution, and no solution. Prove your conjecture. $$\left\\{\begin{array}{l}y=x^{2} \\\y=x+k\end{array}\right.$$

Step-by-Step Solution

Answer

The value of \(k\) for exactly one intersection is \(-\frac{1}{4}\). The system has two solutions for \(k > -\frac{1}{4}\), one solution for \(k = -\frac{1}{4}\), and no solution for \(k < -\frac{1}{4}\).

Step 1: Identify Intersection Points

To find the intersection points of the parabola and the line, set the equations \(y = x^2\) and \(y = x + k\) equal to each other: \(x^2 = x + k\). This leads to the quadratic equation \(x^2 - x - k = 0\).

Step 2: Use the Discriminant

For a quadratic equation in the form \(ax^2 + bx + c = 0\), the discriminant is given by \(b^2 - 4ac\). Here, \(a = 1\), \(b = -1\), and \(c = -k\). Thus, the discriminant is \((-1)^2 - 4(1)(-k) = 1 + 4k\).

Step 3: Determine Number of Solutions

The number of solutions depends on the value of the discriminant: - If \(1 + 4k > 0\), there are two intersection points (two solutions). - If \(1 + 4k = 0\), there is exactly one intersection point (one solution). - If \(1 + 4k < 0\), there are no intersection points (no solution).

Step 4: Solve for Exact Values of k

Set the discriminant equal to zero to find the value of \(k\) for which there is exactly one solution: \[1 + 4k = 0 \]Solving for \(k\) gives: \[k = -\frac{1}{4}\]

Step 5: Make and Prove Conjecture

From the previous steps, you can conjecture: - For \(k > -\frac{1}{4}\), the system has two solutions.- For \(k = -\frac{1}{4}\), the system has one solution. - For \(k < -\frac{1}{4}\), the system has no solution.This is verified by analyzing the discriminant.

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of the form \( ax^2 + bx + c = 0 \). This equation is fundamental when studying the intersection of a parabola, like \( y = x^2 \), and a line, such as \( y = x + k \).
In our specific case, by setting the two equations equal, we derive a new quadratic equation: \( x^2 - x - k = 0 \). This represents a standard form where we have:

• \( a = 1 \)
• \( b = -1 \)
• \( c = -k \)

Deriving this quadratic equation is crucial because it transforms the problem of finding intersection points into a question of solving for the roots of this equation.

Discriminant

The discriminant is a component of the quadratic formula that provides insight into the nature of the roots of a quadratic equation. It is given by \( b^2 - 4ac \). Depending on its value, the discriminant tells us:

• If \( b^2 - 4ac > 0 \), the equation has two distinct real solutions.
• If \( b^2 - 4ac = 0 \), the equation has exactly one real solution.
• If \( b^2 - 4ac < 0 \), there are no real solutions.

In our scenario, substituting \( a = 1 \), \( b = -1 \), and \( c = -k \) into the discriminant gives us \( 1 + 4k \). The value of this expression reveals how many times the line \( y = x + k \) intersects the parabola \( y = x^2 \). For example, if \( 1 + 4k = 0 \), the line is tangent to the parabola, intersecting it exactly once.

Graphing

Graphing is a visual method to understand the behavior of functions and their intersections. In this exercise, graphing the parabola \( y = x^2 \) and various lines \( y = x + k \) on the same coordinate plane can make the concept of intersections more tangible.
As you vary \( k \), the line's position changes:

• If \( k = -\frac{1}{4} \), the discriminant becomes zero, and the line just "kisses" the parabola once, showing one point of intersection.
• If \( k > -\frac{1}{4} \), you observe two intersection points, indicating two solutions.
• If \( k < -\frac{1}{4} \), no intersection points appear, confirming there are no solutions.

A graphing calculator or graph paper can be invaluable tools here, providing a clear picture of how changes in \( k \) affect intersections. This visual approach complements the algebraic method, making it easier to grasp the concept.