Find an equation of the tangent line to the curve at the given point. y = x3 - 2x + 1, (3, 22).
Solution.
The slope at any point of the curve y = x3 - 2x + 1 is defined by its first differential/derivative.
The first derivative of the function y is represented as y’ and is derived as follows:
y' = dy/dx = d(x3 - 2x + 1)/dx = 3x3 - 2
The slope of the given curve at point (3, 22) is y' = 3 × 32 - 2 = 27 - 2 = 25
The slope of the line tangent to the curve is always the same as the slope of the curve at the point at which it is tangent to the curve.
The equation of tangent (straight line) is given by the equation y = mx + b; where ‘m’ is the slope and ‘b’ is a constant.
At the point (3, 22) the value of y is 22 and x is 3.
The value of ‘m’ is obtained from the slope of the curve at the point (3, 22) which is 25 as calculated above.
Therefore the value of the constant can be obtained as follows.
22 = 25 x 3 + b
b = 22 - 75 = -53
Therefore, the equation of the tangent to the curve at the point (3, 22) is y = 25x - 53.
Find an equation of the tangent line to the curve at the given point. y = x3 - 2x + 1, (3, 22).
Summary:
The equation y = x3 - 2x + 1 is a cubic is a polynomial equation. The equation of the tangent to the curve at (3,22) is y = 25x - 53.