Problem 48 - Algebra and Trigonometry

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 \(K^{\prime} s\) 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 point is \( -\frac{1}{4} \).

Step 1: Find Intersection Points

To find where the parabola and the line intersect, we need to solve the system of equations. Set the equations equal: \[ x^2 = x + k \] This simplifies to: \[ x^2 - x - k = 0 \] This quadratic equation will give us the x-values where the parabola and line intersect.

Step 2: Analyze the Discriminant

The number of solutions (intersection points) is determined by the discriminant of the quadratic equation. Recall the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the equation \( x^2 - x - k = 0 \), the discriminant is \( b^2 - 4ac = (-1)^2 - 4(1)(-k) = 1 + 4k \). The discriminant tells us: - Two solutions if \( 1 + 4k > 0 \) - One solution if \( 1 + 4k = 0 \) - No solutions if \( 1 + 4k < 0 \).

Step 3: Solve for k to Find One Intersection Point

Set the discriminant \( 1 + 4k = 0 \) to find when there is exactly one solution:\[ 1 + 4k = 0 \]\[ 4k = -1 \]\[ k = -\frac{1}{4} \] When \( k = -\frac{1}{4} \), the line is tangent to the parabola, and there is exactly one intersection point.

Step 4: Make a Conjecture

Based on the discriminant, we can conjecture the following: - For \( k > -\frac{1}{4} \), there are two intersection points (the parabola and line intersect twice).- For \( k = -\frac{1}{4} \), there is exactly one intersection point (the line is tangent to the parabola).- For \( k < -\frac{1}{4} \), there are no intersection points (the line does not intersect the parabola).

Step 5: Verify the Conjecture

Use a graphing calculator or graphing software to plot the parabola \( y = x^2 \) and lines \( y = x + k \) with different values of \( k \). Check:- For \( k > -\frac{1}{4} \), ensure the line intersects the parabola twice.- For \( k = -\frac{1}{4} \), check the line is tangent to the parabola.- For \( k < -\frac{1}{4} \), confirm there's no intersection.These visual checks will support the conjecture based on the math worked out above.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two, with the general form \( ax^2 + bx + c = 0 \). The coefficients \( a \), \( b \), and \( c \) determine the specific shape and position of the parabola represented by the equation. Importantly, the value of \( a \) influences whether the parabola opens upwards (\( a > 0 \)) or downwards (\( a < 0 \)).
For this exercise, the equation \( y = x^2 \) is a quadratic equation where the parabola opens upwards. Understanding the basic structure of these equations helps in identifying how they might intersect with other graphs, such as a line. Since we are also dealing with linear equations like \( y = x + k \), recognizing the structure aids in setting up simultaneous equations to find their intersection points.

Discriminant

The discriminant is a key component in determining how many solutions a quadratic equation has. It is derived from the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
The part under the square root sign, \( b^2 - 4ac \), is the discriminant. Its value tells us:

• If the discriminant is positive, there are two distinct real solutions (the parabola intersects the line at two points).
• If the discriminant is zero, there is exactly one real solution (the line is tangent to the parabola).
• If the discriminant is negative, there are no real solutions (the line does not intersect the parabola).

In this exercise, when solving \( x^2 - x - k = 0 \), the discriminant is \( 1 + 4k \). By examining this expression, you can predict how many intersection points exist between the parabola \( y = x^2 \) and the line \( y = x + k \).

Graphing Intersections

Understanding graphing intersections involves plotting the graphs of the equations and visually determining where they intersect. This is crucial for verifying algebraic solutions visually. For the given exercise, you plot the parabola \( y = x^2 \) and several lines \( y = x + k \) to see where they meet.
Use graphing tools or graph paper to draw:

• The parabola, which remains the same, as \( y = x^2 \).
• The linear equations like \( y = x + k \) where different values of \( k \) shift the line up or down.

Observe how the number of intersections changes as \( k \) changes. For example, with \( k = -\frac{1}{4} \), the line touches the parabola at exactly one point, highlighting the importance of the discriminant being zero.

Tangent Lines

A tangent line to a curve touches the curve at precisely one point. It does not cross the curve at that point. In the context of this exercise, the line \( y = x + k \) becomes tangent to the parabola \( y = x^2 \) when \( k = -\frac{1}{4} \).
At this value of \( k \), the discriminant \( 1 + 4k \) is zero, confirming a single point of contact. Determining when a line is tangent involves calculating and setting the discriminant to zero in the quadratic equation derived from equating the line and the parabola's equations. Once the line is identified as tangent, it will have exactly one intersection point, visually confirming your algebraic solution. Tangent lines are crucial in calculus and geometry, where they represent the instantaneous rate of change and provide a linear approximation of the curve near the point of tangency.