Equations of Tangents and Normals As mentioned before, the main use for differentiation is to find the gradient of a function at any point on the graph. Having found the gradient at a specific point we can use our coordinate geometry skills to find the equation of the tangent to the curve. To do this we:1. Differentiate the function.2. Put in the x-value intoto find the gradient of the tangent. 3. Put in the x-value into the function (y = ...) to find the coordinates of the point where the tangent touches the curve. 4. Put these values into the formula for a straight line: y - y1 = m (x - x1)where m = gradient and (x1, y1) is where the tangent meets the curve. Example: Find the equation of the tangent to the curve y = x2 - 2x - 3, when x = -1: so when x = -1, and y = 1 + 2 - 3 = 0Therefore the equation of the tangent is y − 0 = -4(x + 1)So now we know that y = -4x - 4 is the equation of the tangent at (-1, 0). The normal to a curve is the line at right angles to the curve at a particular point. This means that the normal is perpendicular to the tangent and therefore the gradient of the normal is -1 × the gradient of the tangent. To find the equation of the normal, follow the same procedures as before, (remembering to multiply the gradient of the tangent by -1 to calculate the gradient of the normal). Example: Find the equation of the normal to the curve y = x3 + x − 10 when x = 2. and y = 8 + 2 − 10 = 0As the gradient of the tangent = 13, the gradient of the normal = -1/13Therefore the equation of the normal is: y − 0 = -(x - 2) / 13 Therefore: 13y = 2 − x is the equation of the normal at (2, 0). Stationary PointsAs mentioned before, the main use for differentiation is to find the gradient of a function at any point on the graph. In particular differentiation is useful to find one of the main features of the graph - the Stationary Points. These are points where the gradient = 0. There are three types of stationary point: There are three possible ways to determine the nature of a stationary point.1. From experience - if you know the shape of the graph, then you know if it is a max/min. All quadratics where the co-efficient of x2 is positive have a minimum (∪ - shaped); all quadratics where the co-efficient of x2 is negative have a maximum (∩ - shaped).2. By looking at the gradient either side of the stationary point. 3. By using the second derivative, which often shown as For a particular value for x, when An example using all three methods: Find the coordinates and nature of the turning points of the curve y = x3 − 12x + 2Firstly, where are the stationary points? Find where the gradient = 0. Therefore: when x = 2 (and y = -14) or when x = -2 (and y = 18). Method 1: We know that a + x3 graph has a maximum followed by a minimum, so (-2, 18) must be a maximum, and (2, -14) must be a minimum. (Also the value of the y-coordinates confirms that this must be true.)Method 2: For this graph the gradient = 0 when x = -2 and x = 2. We can use the fact that the gradient is a multiple of (x + 2)(x - 2) to determine the sign of the gradient either side of these values. This is best illustrated in a table. At x = -2, the gradient goes from positive to negative. This is a ∩-shape, and means that there is a maximum at (-2, 18). At x = 2, the gradient goes from negative to positive. This is a ∪-shape, and means that there is a minimum at (2, -14). Method 3: When x = -2, Therefore there is a maximum at (-2, 18). When x = 2, Therefore there is a minimum at (2, -14). As a guideline - if you know the graph use method 1. Otherwise use method 3 unless the second derivative is hard to find. If you do not know the shape of the graph, or you cannot differentiate twice or then use the method 2.