Question

equation of the tangent at ​

Answer 1

Answer:

1) Find the first derivative of f(x). 2) Plug x value of the indicated point into f '(x) to find the slope at x. 3) Plug x value into f(x) to find the y coordinate of the tangent point. 4) Combine the slope from step 2 and point from step 3 using the point-slope formula to find the equation for the tangent line.

Step-by-step explanation:

Remember, if two lines are perpendicular, the product of their gradients is -1. So if the gradient of the tangent at the point (2, 8) of the curve y = x3 is 12, the gradient of the normal is -1/12, since -1/12 × 12 = -1 . hence the equation of the normal at (2,8) is 12y + x = 98 .

A tangent to a circle at point P with coordinates is a straight line that touches the circle at P. The tangent is perpendicular to the radius which joins the centre of the circle to the point P. As the tangent is a straight line, the equation of the tangent will be of the form y = m x + c .