Question

Find the equation of the normal to the curve \( y = x^2 - x \) at the coordinate point position \( (1, 0) \).

• A. \( x + y - 1 = 0 \)
• B. \( x - y - 1 = 0 \)
• C. \( x + y + 1 = 0 \)
• D. \( 2x + y - 2 = 0 \)

Hint:

Always read the question carefully to see if it asks for the equation of the tangent or the normal. Forgetting to invert and negate the slope for a normal line is a common oversight under exam time pressure.

Answer 1

The Correct Option is A

Concept: The slope of the tangent line (\( m_t \)) to a curve at a given point is found by calculating its first derivative at that point. Because a normal line runs perpendicular to the tangent, its slope (\( m_n \)) is the negative reciprocal of the tangent slope: \[ m_n = -\frac{1}{m_t} = -\frac{1}{\left(\frac{dy}{dx}\right)} \]

Step 1: Calculate the first derivative and find the tangent slope.
Differentiate the curve's equation with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx}\left( x^2 - x \right) = 2x - 1 \] Evaluate this derivative at the target x-coordinate (\( x = 1 \)) to find the tangent slope: \[ m_t = 2(1) - 1 = 1 \]

Step 2: Calculate the perpendicular slope of the normal line.
Take the negative reciprocal of the tangent slope: \[ m_n = -\frac{1}{1} = -1 \]

Step 3: Set up the line equation using point-slope form.
Using the point coordinates \( (1,0) \) and our normal slope \( m_n = -1 \), write the line equation: \[ y - y_1 = m_n(x - x_1) \quad \Rightarrow \quad y - 0 = -1(x - 1) \] Distribute the negative sign and rearrange all terms to the left side: \[ y = -x + 1 \quad \Rightarrow \quad x + y - 1 = 0 \]