Problem 48 - Precalculus Mathematics for Calculus

Question

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 system has one solution when \( k = -\frac{1}{4} \). For \( k > -\frac{1}{4} \), two solutions, and for \( k < -\frac{1}{4} \), no solutions.

Step 1: Graph the Parabola

To start, plot the parabola given by the equation \( y = x^2 \) on graph paper or using a graphing calculator. The parabola opens upwards with its vertex at the origin \((0, 0)\).

Step 2: Graph Linear Equations

Next, graph the linear equation \( y = x + k \) for various values of \( k \). Choose values such as \( k = -3, -1, 0, 1, 3 \) to see how the line shifts vertically on the graph.

Step 3: Identify Intersection Points

For each value of \( k \), observe how many points the line intersects with the parabola. When \( y = x + k \) intersects \( y = x^2 \) at two points, the values of \( k \) result in two solutions. Adjust \( k \) to find when the line is tangent to the parabola, yielding exactly one solution.

Step 4: Find Value of k for Tangency

Set the two equations equal to find when they intersect: \( x^2 = x + k \). Rearrange this into a quadratic equation: \( x^2 - x - k = 0 \). Use the discriminant \( b^2 - 4ac \) from the quadratic formula to determine the number of solutions. The quadratic has exactly one solution when the discriminant is zero: \( (-1)^2 - 4(1)(-k) = 0 \).

Step 5: Solve for k

Solve \( 1 + 4k = 0 \) to find \( k = -\frac{1}{4} \). This is the value for which there is exactly one intersection point, meaning the line is tangent to the parabola.

Step 6: Formulate Conjecture and Proof

Conjecture: The system \( y = x^2 \) and \( y = x + k \) has two solutions if \( k > -\frac{1}{4} \), one solution if \( k = -\frac{1}{4} \), and no solutions if \( k < -\frac{1}{4} \). Proof: This is confirmed by analyzing the discriminant \( 1 + 4k \), which indicates two positive roots when \( 1 + 4k > 0 \), one root when \( 1 + 4k = 0 \), and no roots when \( 1 + 4k < 0 \).

Key Concepts

Graphing Parabolas

Visualization is a powerful tool when dealing with algebraic equations, and graphing is an excellent way to gain insights into these equations. A parabola is a symmetrical curve, and the simplest form of a parabola is given by the equation \(y = x^2\). To graph this parabola, plot points for various values of \(x\).

• The vertex is located at the origin \((0, 0)\), which is the lowest point on the curve for \(y = x^2\).
• This parabola opens upwards because the coefficient of \(x^2\) is positive.
• The graph is symmetrical along the \(y\)-axis. This means if you fold the graph along the \(y\)-axis, the two halves would match perfectly.

By sketching the curve, you can visually analyze how other equations, such as lines, relate to the parabola. Understanding the characteristics of this curve helps with more complex topics like finding intersection points.

Linear Equations

Linear equations in graphing represent straight lines. The equation \(y = x + k\) is a linear equation where \(k\) determines the vertical shift of the line. By varying \(k\) you can draw multiple parallel lines on the coordinate plane:

• For each different \(k\) value, the line \(y = x + k\) moves up or down, without changing its slope, which is 1 (since it follows the form \(y = mx + c\) where \(m=1\)).
• For instance, if \(k = 0\), the line \(y = x\) passes through the origin. If \(k = 2\), the line is shifted two units up.
• This linear equation allows for flexibility in visualizing how changes in \(k\) affect its position relative to the parabola \(y = x^2\).

Exploring different \(k\) values helps understand how lines can intersect, be tangent, or not meet a parabola at all.

Intersection Points

Intersection points occur where a line and parabola have the same \(x\) and \(y\) coordinates. Mathematically, it means solving for \(x\) in the system \(\begin{array}{l} y = x^2 \ y = x + k \end{array}\). By equating \(x^2 = x + k\), you obtain a quadratic equation: \(x^2 - x - k = 0\).

• If this quadratic equation yields two solutions for \(x\), there are two intersection points.
• If it has one solution, there's one intersection point (tangency).
• No solutions indicate no intersection.

Analyzing when these intersections occur involves using concepts like the discriminant from the quadratic formula, which lets you determine the number of roots (solutions) of the quadratic equation.

Tangent Lines

A tangent line to a curve is a line that touches the curve at exactly one point. In the context of our parabola \(y = x^2\) and linear equation \(y = x + k\), determining when they are tangent involves setting \(x^2 = x + k\) such that the quadratic equation \(x^2 - x - k = 0\) has exactly one solution. This happens when its discriminant \(b^2 - 4ac\) is zero.

• For \(x^2 - x - k = 0\), the discriminant \((-1)^2 - 4(1)(-k) = 1 + 4k\) determines the nature of solutions.
• Setting \(1 + 4k = 0\) and solving for \(k\), we get \(k = -\frac{1}{4}\). This is the value where the line is tangent to the parabola.
• Tangency implies that at \(x\), \(y = x^2\) equals \(y = x + k\) for only one value, indicating a special point of touch rather than crossing.

Identifying tangent lines is crucial for understanding deeper geometric relationships in curves and lines.