To find we must find a point on the line, namely the point on the curve when = 2. c x When = 2 the value on the curve is 3 22 7 2 5 3 ,      x y i.e. when = 2, = 3. x y Substituting these values in = 5x + we get 3 = 5 2 + = y c  c  c –7  the equation of the tangent is = 5x 7. y – Differentiation from 1st principles Example Stationary points A stationary point is when the gradient to a curve is zero. The gradient of a curve at any point is found by differentiating. Hence to find a stationary point differentiate (to get gradient) and equate it to zero. Example Find expression for gradient by differentiating Stationary point when gradient is zero so Find coordinate from original equation: When Coordinates are (0,11) and (2,7)
To find c we must find a point on the line, namely the point on the curve when x = 2. When x = 2 the y value on the curve is 3  22  7  2  5  3 , i.e. when x = 2, y = 3. Substituting these values in y = 5x + c we get 3 = 5  2 + c  c = –7  the equation of the tangent is y = 5x – 7. Differentiation from 1st principles Example Stationary points A stationary point is when the gradient to a curve is zero. The gradient of a curve at any point is found by differentiating. Hence to find a stationary point differentiate (to get gradient) and equate it to zero. Example Find expression for gradient by differentiating Stationary point when gradient is zero so Find coordinate from original equation: When Coordinates are (0,11) and (2,7)