How do you find the equation of the line tangent to #f(x)= x^3#, at (2,8)?

One form of the equation of a straight line is y = mx + c , where m, represents the gradient and c , the y-intercept. To find m, require to obtain f'(x) and the value of f'(2).

equation is then : y = 12x + c

use (2 , 8 ) in equation to find c.

hence : 8 = 12(2) + c → c = 8 - 24 = - 16

To find the equation of the line tangent to the function f(x) = x^3 at the point (2,8), we can use the concept of differentiation.

First, we need to find the derivative of the function f(x) = x^3. The derivative of x^n (where n is a constant) is given by nx^(n-1). Applying this rule, the derivative of f(x) = x^3 is f'(x) = 3x^2.

Next, we substitute the x-coordinate of the given point (2,8) into the derivative to find the slope of the tangent line. Plugging x = 2 into f'(x) = 3x^2, we get f'(2) = 3(2)^2 = 12.

Now that we have the slope of the tangent line, we can use the point-slope form of a linear equation to find the equation of the line. The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.

Substituting the values (2,8) and m = 12 into the point-slope form, we get the equation of the tangent line as y - 8 = 12(x - 2).

Simplifying this equation, we can rewrite it in slope-intercept form (y = mx + b) as y = 12x - 16.

Therefore, the equation of the line tangent to f(x) = x^3 at the point (2,8) is y = 12x - 16.