find the equation of a curve whose gradient at point coordinate (4,2) is -0.666666

To find the equation of the curve, we need to determine the slope of the tangent line at any point on the curve. In this case, we are given that the gradient at the point (4,2) is -0.666666.

Let's assume the equation of the curve is y = f(x). The slope of the tangent line at any point (x, f(x)) on the curve is given by the derivative of f(x) with respect to x.

The derivative of f(x) is denoted as f'(x) or dy/dx. Given that the gradient at the point (4,2) is -0.666666, we can write:

f'(4) = -0.666666

Now, let's find the equation of the curve by integrating f'(x) with respect to x.

$$\int f'(x) \, dx = \int -0.666666 \, dx$$

Integrating both sides:

$$f(x) = -0.666666x + C$$

where C is a constant of integration.

To find the value of C, we can use the point (4,2) that lies on the curve. Plugging in the values:

2 = -0.666666(4) + C

2 = -2.666664 + C

C = 2 + 2.666666

C = 4.666666

Therefore, the equation of the curve is:

$$f(x) = -0.666666x + 4.666666$$

Answer: The equation of the curve is y = -0.666666x + 4.666666.