C1: Tangents and Normals Learning Objective: to find the equation of a tangent and a normal to a curve at a given point by applying the rules of differentiation

Starter: Differentiate y = 2x4 + 7x3 - 5x2 y = ½ x4 y = 2√x y = (2x8 - 3x4 +5x2)/ x3

Tangents and normals Remember, the tangent to a curve at a point is a straight line that just touches the curve at that point. The normal to a curve at a point is a straight line that is perpendicular to the tangent at that point. You may wish to ask students how we can verify that the curve y = x2 – 5x + 8 passes through the point (3, 2). Substituting x = 3 into y = x2 – 5x + 8 gives y = 2 as required. We can use differentiation to find the equation of the tangent or the normal to a curve at a given point. For example: Find the equation of the tangent and the normal to the curve y = x2 – 5x + 8 at the point P(3, 2).

Tangents and normals y = x2 – 5x + 8 At the point P(3, 2) x = 3 so: 4 The gradient of the tangent at P is therefore 4. Using y – y1 = m(x – x1) give the equation of the tangent at the point P(3, 2): y – 2 = 4(x – 3) y – 2 = 4x – 12 y – 4x + 10 = 0

Tangents and normals The normal to the curve at the point P(3, 2) is perpendicular to the tangent at that point. The gradient of the tangent at P is 4 and so the gradient of the normal is Using y – y1 = m(x – x1) give the equation of the tangent at the point P(3, 2):

Task 1 Find the equation of the tangent to the curve y = x2 - 7x + 10 at the point (2, 0). Find the equation of the normal to the curve y = x2 – 5x at the point (6, 6). Find the equations of the normals to the curve y = x + x3 at the points (0, 0) and (1, 2), and find the co-ordinates of the point where these normals meet. For f(x) = 12 – 4x + 2x2, find an equation of the tangent and normal, at the point where x = -1 on the curve with equation y = f(x).