Question

The gradient of a curve at any point p(x,y) on it is given by (4x-3x²+1). If the curve passes through the point A(2,1). Find:

a. The equation of the curve
b. The equation of the tangent of the curve at A

a. y = x³ - x⁴ + x + C; b. y = 3x - 1
a. y = x³ - 2x² + x + C; b. y = 3x - 1
a. y = 2x³ - x² + x + C; b. y = 3x - 1
a. y = x³ - 3x² + x + C; b. y = 3x - 1

Answer 1

The equation of the curve is y = x³ - 2x² + x + C, after integration and applying the point A(2,1) to find C. The equation of the tangent at point A(2,1) is y = 3x - 5, with 3 being the gradient of the curve at x=2.

Step-by-step explanation:

To find the equation of the curve, we integrate the given gradient function (4x-3x²+1). The indefinite integral of the gradient function gives us the equation of the curve, which, after integration, is y = x³ - x´ + x + C. To find the constant C, we use the fact that the curve passes through the point A(2,1), leading us to the equation y = x³ - 2x² + x + C. Substituting x = 2 and y = 1 gives us the specific value of C.

For the equation of the tangent at point A, we use the gradient of the curve at x = 2, given by 3, to write the equation of the tangent in the form y = mx + b, where m is the gradient. The point (2,1) lies on this tangent, so by substituting, we find the y-intercept b. The equation of the tangent at A is y = 3x - 5.