How to Find an Equation for the Line Perpendicular to the Tangent Line to the Curve y = x³ + 4x + 1 at the Point (2, 1)

To find an equation for the line perpendicular to the tangent line to the given curve at the specified point, we can follow these steps:

1. Find the derivative of the curve: The first step is to calculate the derivative of the curve, which will give us the slope of the tangent line at any point on the curve. For the function y = x³ + 4x + 1, the derivative is:
• dy/dx = 3x² + 4
2. Evaluate the derivative at the point (2, 1): Next, we will find the slope of the tangent line at the point (2, 1) by substituting x = 2 into the derivative:
• dy/dx (2) = 3(2)² + 4 = 3(4) + 4 = 12 + 4 = 16
3. Determine the slope of the perpendicular line: The slope of the line that is perpendicular to the tangent line is the negative reciprocal of the tangent line’s slope. Thus, if the slope of the tangent line is 16, the slope of the perpendicular line (m) is:
• m = -1/16
4. Use point-slope form to find the equation: Now that we have the slope of the perpendicular line, we can use the point-slope form of a line’s equation, which is expressed as:
• y – y₁ = m(x – x₁)

Substituting the slope (-1/16) and the coordinates of the point (2, 1):

• y – 1 = -1/16(x – 2)
5. Simplifying the equation: Now we can simplify this equation to express it in slope-intercept form (y = mx + b):
• Distributing the slope: y – 1 = -1/16x + 2/16
• Adding 1 to both sides: y = -1/16x + 2/16 + 16/16
• Combining constants: y = -1/16x + 18/16

Finally, we can express this as:

y = -1/16x + 9/8

This gives us the equation of the line that is perpendicular to the tangent line at the point (2, 1) on the curve.