Problem 52 - Calculus Early Transcendentals

Question

Find \(k\) if the curve \(y=x^{2}+k\) is tangent to the line \(y=2 x\)

Step-by-Step Solution

Answer

The value of \(k\) is 1.

Step 1: Understand the Problem

We need to find the value of \(k\) such that the curve \(y = x^2 + k\) is tangent to the line \(y = 2x\). For tangency, the curve and the line must meet at exactly one point, and their slopes must be equal at that point.

Step 2: Set the Equations Equal

For the curve and the line to intersect, set \(x^2 + k = 2x\). Rearrange this to obtain a quadratic equation: \(x^2 - 2x + k = 0\).

Step 3: Use the Tangency Condition

A quadratic equation has exactly one solution (tangent) when its discriminant is zero. The discriminant \(D\) of the equation \(ax^2 + bx + c = 0\) is given by \(b^2 - 4ac\). For our equation, \(a = 1\), \(b = -2\), \(c = k\).

Step 4: Set the Discriminant to Zero

Substitute the values of \(a\), \(b\), and \(c\) into the discriminant formula: \((-2)^2 - 4(1)(k) = 0\). Simplify this to: \(4 - 4k = 0\).

Step 5: Solve for \(k\)

Rearrange the equation \(4 - 4k = 0\) to find \(k\). This gives \(4 = 4k\). Divide both sides by 4 to obtain \(k = 1\).

Key Concepts

Tangency

Tangency in mathematics refers to the point where a curve and a line touch, but do not cross. This concept is significant because, at the tangent point, the curve and the line share the same direction, which means their slopes are equal.
This condition helps to solve real-world problems, such as determining the maximum or minimum values in a given scenario.

• For two functions to be tangent, they must intersect at exactly one point.
• At the tangency point, the derivative (slope) of the curve will equal the slope of the line.

This was applied in our original exercise by ensuring the line and curve had the same slope where they met, confirming they touched but did not cross.

Quadratic Equations

Quadratic equations are polynomials of degree two, in the form of \(ax^2 + bx + c = 0\). These equations graph as a parabola and can have two solutions, one solution, or no real solution depending on the discriminant.

• Solutions of a quadratic equation can be found using the quadratic formula:
  \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
• The vertex of the parabola gives either a maximum or minimum point of the curve.

In our exercise, the equation \(x^2 - 2x + k = 0\) is in this quadratic form. Finding \(k\) ensures the parabola \(y = x^2 + k\) is tangent to the line \(y = 2x\), by using properties of quadratic equations and tangency.

Discriminant

The discriminant of a quadratic equation, \(b^2 - 4ac\), is a tool that provides significant information about the nature of the roots of the equation.

• If the discriminant is positive, the equation has two distinct real solutions.
• If it is zero, there is exactly one real solution, indicating a tangency point where two graphs touch once.
• If the discriminant is negative, no real solutions exist, suggesting that the graph does not cross the x-axis.

In the given problem, setting the discriminant to zero enabled us to find the value of \(k\). This indicated the curve \(y = x^2 + k\) was precisely tangent to the line \(y = 2x\) at a single unique point. In this case, the discriminant \((4 - 4k = 0)\) revealed \(k = 1\), resulting in one shared point of intersection.