Problem 52 - Precalculus: Mathematics 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 line intersects the parabola at exactly one point when \( k = -\frac{1}{4} \). For \( k > -\frac{1}{4} \), there are two intersection points. For \( k < -\frac{1}{4} \), there are no intersection points.

Step 1: Find the Intersection Points Algebraically

To find where the line and the parabola intersect, set the equations equal to each other: \[ x^2 = x + k \]Rearrange it to form a quadratic equation:\[ x^2 - x - k = 0 \]

Step 2: Analyze the Discriminant of the Quadratic

For a quadratic equation \( ax^2 + bx + c = 0 \) to have one solution, the discriminant \( b^2 - 4ac \) must be zero. For our equation \( x^2 - x - k = 0 \), we have:\[ a = 1, \, b = -1, \, c = -k \]So the discriminant is:\[ (-1)^2 - 4(1)(-k) = 1 + 4k \]

Step 3: Solve for One Intersection Point

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

Step 4: Make a Conjecture

For the quadratic equation to have two solutions (i.e., the line intersects the parabola at two points), the discriminant must be positive: \( 1 + 4k > 0 \). Hence, \( k > -\frac{1}{4} \).For no solutions (i.e., no intersection), the discriminant must be negative: \( 1 + 4k < 0 \). Therefore, \( k < -\frac{1}{4} \).

Step 5: Prove the Conjecture

Based on the discriminant, when \( k = -\frac{1}{4} \), \( 1 + 4k = 0 \), hence there is exactly one solution. If \( k > -\frac{1}{4} \), the discriminant is positive, so the equation \( x^2 - x - k = 0 \) has two distinct solutions, implying the line intersects the parabola at two points. If \( k < -\frac{1}{4} \), the discriminant is negative, and there is no real solution, meaning the line and parabola do not intersect.

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of the second degree. It is generally represented in the form:\[ ax^2 + bx + c = 0 \]where:

• \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \)
• \( x \) represents the variable

The shape of the graph of a quadratic equation is a parabola, and its orientation (opening upwards or downwards) is determined by the sign of \( a \). In our problem, the intersection points of the parabola \( y = x^2 \) and the line \( y = x + k \) can be found by equating their equations and solving the resulting quadratic equation, \( x^2 - x - k = 0 \). This equation helps us discover the points where the shapes meet or intersect one another on the grid.
Thus, understanding the basic form of a quadratic is essential to solving intersection problems.

Discriminant

The discriminant is a special value derived from the coefficients of a quadratic equation and is crucial for determining the nature of the roots (solutions) of the equation. The discriminant \( D \) is given by the formula: \[ D = b^2 - 4ac \]where \( a \), \( b \), and \( c \) are coefficients from the quadratic equation \( ax^2 + bx + c = 0 \).
It provides insight into the number of solutions the quadratic equation has:

• If \( D > 0 \), there are two distinct real roots.
• If \( D = 0 \), there is exactly one real root, also known as a double root.
• If \( D < 0 \), there are no real roots, implying the solutions are complex or imaginary.

For our quadratic \( x^2 - x - k = 0 \), the discriminant \( D \) is \( 1 + 4k \), and analyzing its value helps determine how the line interacts with the parabola.

Graphical Solutions

Graphical solutions involve plotting equations on a coordinate plane to find where they intersect. This visual representation can offer a quick and intuitive understanding of how solutions relate. To solve for the intersection of the line \( y = x + k \) and the parabola \( y = x^2 \) graphically, you would:

• Draw the parabola \( y = x^2 \), which has a vertex at the origin \((0, 0)\) and opens upwards.
• Draw the line \( y = x + k \) for various \( k \) values to visualize different intersection scenarios.
• Observe where the line and parabola meet on the graph. Each point of intersection represents a real solution or root of the quadratic equation.

Graphing complements algebraic methods by providing visual confirmation of the number and position of solutions.
This approach is especially helpful when predicting the behavior of intersections as parameters like \( k \) change.

Parabola-Linear Intersection

The intersection of a parabola and a line is a common problem in mathematics, crucial for understanding quadratic functions and their applications. Finding these intersections helps us understand behavior such as:

• Where paths meet or objects trajectories intersect.
• Solving real-world problems involving quadratic relationships.

In our scenario, the line \( y = x + k \) and the parabola \( y = x^2 \) intersect when solving the quadratic equation \( x^2 - x - k = 0 \). The number of intersections (solutions) depends on the discriminant value:

• For \( k = -\frac{1}{4} \), there is exactly one intersection, as confirmed by setting the discriminant \( 1 + 4k \) to zero.
• For \( k > -\frac{1}{4} \), there are two intersections, as the discriminant is positive.
• For \( k < -\frac{1}{4} \), there are no intersections, indicating that the discriminant is negative.

Parabola-linear intersections illustrate the real-life applications of quadratic functions and linear equations.