Question

There are two tangent lines to the curve \(y=4 x-x^{2}\) that go through \((2,5)\). Find the equations of both of them. Hint: Let \(\left(x_{0}, y_{0}\right)\) be a point of tangency. Find two conditions that \(\left(x_{0}, y_{0}\right)\) must satisfy. See Figure 4 .

Short answer

The tangent line equations are \(y = (2 - 2\sqrt{2})x + (9 - 4\sqrt{2})\) and \(y = (2 + 2\sqrt{2})x + (9 + 4\sqrt{2})\)."

Step-by-step solution

Identify the Point of Tangency

First, denote the point of tangency on the curve as \((x_0, y_0)\). Since this point lies on the curve \(y = 4x - x^2\), its coordinates satisfy the equation of the curve. Therefore, \(y_0 = 4x_0 - x_0^2\).

Determine the Slope of the Tangent Line

The derivative of the curve, \(y = 4x - x^2\), gives the slope of the tangent line at any point \(x\). Compute the derivative, \(y' = 4 - 2x\). At the point of tangency \((x_0, y_0)\), the slope is \(4 - 2x_0\).

Equation of the Tangent Line at the Point of Tangency

The equation of the tangent line at the point \((x_0, y_0)\) with slope \(m = 4 - 2x_0\) is given by the point-slope form: \(y - y_0 = (4 - 2x_0)(x - x_0)\).

Equation Condition for Passing through (2,5)

Substitute \((2,5)\) into the tangent line equation: \(5 - y_0 = (4 - 2x_0)(2 - x_0)\). This results in the equation \(5 - (4x_0 - x_0^2) = (4 - 2x_0)(2 - x_0)\).

Solve for x_0

Simplify and solve the equation \(5 - (4x_0 - x_0^2) = (4 - 2x_0)(2 - x_0)\): 1. Expand the right-hand side: \((4 - 2x_0)(2 - x_0) = 8 - 4x_0 - 2x_0 + 2x_0^2\).2. Simplify the equation: \(5 - 4x_0 + x_0^2 = 8 - 6x_0 + 2x_0^2\).3. Rearrange terms: \(x_0^2 - 2x_0 + 3 = 0\).4. Solve using the quadratic formula: \(x_0 = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 3}}{2}\) which simplifies to \(x_0 = 1 \pm \sqrt{2}\).

Find y_0 for each x_0

Substitute \(x_0 = 1 + \sqrt{2}\) and \(x_0 = 1 - \sqrt{2}\) back into the curve equation \(y_0 = 4x_0 - x_0^2\) to find the corresponding \(y_0\) for each tangent point.

Write the Equations of Tangent Lines

Using both points of tangency \((1 + \sqrt{2}, y_0)\) and \((1 - \sqrt{2}, y_0)\), and the slopes \(2(1 - \sqrt{2})\) and \(2(1 + \sqrt{2})\), write the full equation of each tangent line.

Key concepts

Derivative

Understanding derivatives is crucial when dealing with tangent lines. A derivative represents the rate of change of a function.
In mathematical terms, it provides the slope of the tangent line at any given point on a curve. When we say we want to find the equation of a tangent line, we're talking about finding a line that touches the curve at exactly one point, with a slope equal to the derivative of the curve at that point.
In the exercise, the curve is defined by the equation \(y = 4x - x^2\), and its derivative is found by differentiating with respect to \(x\). Doing so, we get \(y' = 4 - 2x\). This derivative \(y'\) gives the slope of the tangent line at any point \((x_0, y_0)\) on the curve.

• Finding the derivative is like bringing a magnifying glass to a curve to figure out exactly how it bends at each point.
• The derivative provides the critical "tilt" or inclination of the tangent line, indicating whether the curve is rising or falling as it passes through that point.

Equation of a Line

The equation of a line provides a mathematical description of that line in the coordinate plane.
Typically, we express this equation in one of several forms, with the point-slope form being highly useful for our purposes: \(y - y_0 = m(x - x_0)\). Here, \((x_0, y_0)\) is a known point on the line, and \(m\) is the slope.
For a tangent line, this form is instrumental because if you already know the point of tangency \((x_0, y_0)\) and have calculated the slope from the derivative, you can directly write the equation of the tangent line.

• The point-slope equation shows directly how every point on the line relates to that initial point of tangency.
• It's a direct path from understanding the slope at a point to describing the entire line.

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\).
This formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), provides the roots or solutions for the equation.
In our exercise, after setting up the equation satisfied by the points of tangency, the problem simplifies to \(x_0^2 - 2x_0 + 3 = 0\). The coefficient \(a = 1\), \(b = -2\), and \(c = 3\) allow you to substitute these into the quadratic formula, solving for \(x_0\). This helps us find the specific \(x\)-coordinates of the points where the tangents touch the curve.

• The quadratic formula is like a key that unlocks the precise points where equations intersect the axis.
• It gives us a way to handle equations that aren't immediately solvable by looking at them.

Point of Tangency

The point of tangency is the specific point where a tangent line touches a curve and does not cross it.
This point is crucial in determining both the location and the characteristics of the tangent line.
Determining the point of tangency involves understanding both geometry and calculus. You need to find where the tangent line's slope, provided by the derivative, aligns exactly with the curve while still sharing a common point.
For the problem, points of tangency are found after establishing and solving the equations derived using the quadratic formula for \(x_0\). With \(x_0 = 1 \pm \sqrt{2}\), these solutions define the \(x\)-coordinates of the points.

• The point of tangency is the meeting point where algebra meets geometry in a single spot on the curve.
• Recognizing this point gives clarity on the characteristics of the curve at that very spot, like how fast it's rising or falling.