Lessons 6–7 · 2 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 11, Parts 1 and 2, 1st edition. ZAMIN NASHR, Tashkent, 2018
6
Equations of the tangent and the normal to a graph
Textbook, Part 1: pp. 34–38
GoalWrite the equations of the tangent and the normal to a graph at a given point and find a tangent parallel to a given line.
New words
tangent equation · urinma tenglamasinormal line · normalcondition of parallelism · parallellik sharticondition of perpendicularity · perpendikularlik sharti
Explanation
The tangent to y = f(x) at the point with abscissa x₀ has slope f′(x₀) and passes through (x₀, f(x₀)), so its equation is y − f(x₀) = f′(x₀)(x − x₀). The normal is the line perpendicular to the tangent at that point; since the product of the slopes is −1, for f′(x₀) ≠ 0 its equation is y − f(x₀) = −(1/f′(x₀))(x − x₀). If f′(x₀) = 0 the tangent is horizontal (y = f(x₀)) and the normal is vertical (x = x₀). To find a tangent parallel to y = kx + b, solve f′(x₀) = k for x₀ and then write the tangent equation. For the angle α between the tangent and the positive Ox direction, tan α = f′(x₀).
Worked examples
f(x) = x³ − 3x, x₀ = 2. f(2) = 2, f′(x) = 3x² − 3, f′(2) = 9. Tangent: y − 2 = 9(x − 2), i.e. y = 9x − 16. Normal: y − 2 = −(1/9)(x − 2), i.e. y = −x/9 + 20/9.
Tangents to f(x) = x³ − 2x parallel to y = x + 4: f′(x) = 3x² − 2 = 1 ⇒ x₀ = ±1. f(1) = −1: y + 1 = x − 1, i.e. y = x − 2; f(−1) = 1: y − 1 = x + 1, i.e. y = x + 2. A horizontal tangent, for example for y = x² − 4x + 5: 2x − 4 = 0 ⇒ x₀ = 2, f(2) = 1, equation y = 1.
Class activity
“Two lines”: each pair picks a function (say y = x² − 2x) and a point on its graph, writes the tangent and normal equations and draws them on a coordinate grid. Another pair checks with a protractor or set square that the drawing shows a right angle.
Practice
1
Find the slope of the tangent to the graph of f(x) = x² + 2x at x₀ = −3.
-4
2
Write the equation of the tangent to y = eˣ at x₀ = 0.
f(0) = 1, f′(0) = 1: y − 1 = x ⇒ y = x + 1.
3
At which points of the graph of y = x³ − 3x is the tangent horizontal?
3x² − 3 = 0 ⇒ x = ±1: at the points (1, −2) and (−1, 2).
4
Why is the slope of the normal −1/f′(x₀), and what is the normal when f′(x₀) = 0?
Perpendicular lines have slopes whose product is −1. When f′(x₀) = 0 the tangent is horizontal, so the perpendicular line is vertical: x = x₀.