Find the equation of the tangent and normal lines to the curve y = x3 - 3x − 1 at the point (1, −3).

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Here are the solutions for parts (a) and (b). Given the curve y = x³ − 3x − 1 and the point (1, −3). Step 1: Find the first derivative of the curve to get the slope of the tangent. dy/dx = d/dx(x³ − 3x − 1) = 3x² − 3 Step 2: Calculate the slope of the tangent at the point (1, −3). Substitute x=1 into the derivative: m_t = dy/dx |_x=1 = 3(1)² − 3 = 3 − 3 = 0 Part (a): Equation of the tangent Step 3: Use the point-slope form y − y₁ = m_t(x − x₁) with m_t = 0 and (x₁, y₁) = (1, −3). y − (−3) = 0(x − 1); y + 3 = 0; y = −3; The equation of the tangent is y = −3. Part (b): Equation of the normal Step 4: Calculate the slope of the normal. The slope of the normal m_n is the negative reciprocal of the slope of the tangent m_t. Since m_t = 0, the tangent line is horizontal. Therefore, the normal line will be vertical. A vertical line has an undefined slope and its equation is of the form x = c. Since the normal passes through (1, −3), its x-coordinate must be 1. The equation of the normal is x = 1.

Point	x	y
x-intercept	−1.53	0
Maximum	−1	1
x-intercept	−0.35	0
y-intercept	0	−1
Minimum	1	−3
x-intercept	1.88	0

Find the equation of the tangent and normal lines to the curve y = x3 − 3x − 1 at the point (1, -3).

Answers

Step-by-step solution

Part a· 1 step

Use the point-slope form $y - y_1 = m_t(x - x_1)$ with $m_t = 0$ and $(x_1, y_1) = (1, -3)$ .

Part b· 1 step
$x = 1$

Calculate the slope of the normal.

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