By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
A tangent line to a curve at a given point is a line that just touches the curve at that point. The normal line, on the other hand, is a line perpendicular to the tangent line at that point. Finding the equations of tangent and normal lines is crucial in various fields, including physics, engineering, and economics.
Tangent and normal lines appear in real-world applications such as:
Find the equation of the tangent line to the function $f(x) = x^2 + 3x - 2$ at the point $(1, 4)$.
$$ \begin{aligned} f'(x) &= 2x + 3 \ f'(1) &= 2(1) + 3 = 5 \ \text{Equation of tangent line: } y - 4 &= 5(x - 1) \ y &= 5x - 1 \end{aligned} $$
Find the equation of the normal line to the function $f(x) = x^2 - 2x - 3$ at the point $(-1, 0)$.
$$ \begin{aligned} f'(x) &= 2x - 2 \ f'(-1) &= 2(-1) - 2 = -4 \ \text{Slope of normal line: } m &= -\frac{1}{-4} = \frac{1}{4} \ \text{Equation of normal line: } y - 0 &= \frac{1}{4}(x + 1) \ y &= \frac{1}{4}x + \frac{1}{4} \end{aligned} $$
Find the equation of the tangent line to the function $f(x) = \sin x$ at the point $(0, 0)$.
$$ \begin{aligned} f'(x) &= \cos x \ f'(0) &= \cos 0 = 1 \ \text{Equation of tangent line: } y - 0 &= 1(x - 0) \ y &= x \end{aligned} $$
What is the equation of the tangent line to the function $f(x) = x^2 + 3x - 2$ at the point $(1, 4)$?
A) $y = 5x - 1$ B) $y = 2x + 3$ C) $y = x - 1$ D) $y = x + 1$
The correct answer is A) $y = 5x - 1$ because this is the equation of the tangent line to the function $f(x) = x^2 + 3x - 2$ at the point $(1, 4)$.
What is the slope of the normal line to the function $f(x) = x^2 - 2x - 3$ at the point $(-1, 0)$?
A) $-\frac{1}{4}$ B) $\frac{1}{4}$ C) $-2$ D) $2$
The correct answer is A) $-\frac{1}{4}$ because this is the slope of the normal line to the function $f(x) = x^2 - 2x - 3$ at the point $(-1, 0)$.
What is the equation of the tangent line to the function $f(x) = \sin x$ at the point $(0, 0)$?
A) $y = x$ B) $y = x + 1$ C) $y = x - 1$ D) $y = x^2$
The correct answer is A) $y = x$ because this is the equation of the tangent line to the function $f(x) = \sin x$ at the point $(0, 0)$.
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