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# Equation of a Tangent line

 Finding the equation of a tangent line Differentiate then use the derivative and a given 'x' value or a given point, to find the gradient of the tangent line y = mx + c Substitute in a known 'x' and 'y' value and solve to find 'c' This will determine the equation of the tangent line. Ex15.03, Nulake IAS p84,85

### Example

 Find the equation of the tangent to the curve y = x2 - 3x + 5 at the point (1,3) Steps: y = x2 - 3x + 5 1) General tangent equation y = mx + c 2) Differentiate to find gradient m y' = 2x - 3 3) Substitute x = 1 into the derivative y' = 2x1 - 3 y' = -1 4) The result is the gradient m m = -1 Tangent equation so far is y = -1x + c 5) Substitute in x=1 & y=3 values in to find c y = -1x + c 3 = -1x1 + c 3 = -1 + c 4 = c 6) Tangent equation is y = -1x + 4

# Equation of a Normal Line (extension)

### The normal is a line perpendicular (at 90º) to the tangent line. web linkFind the equation of the normal line of y = 0.5x2 - 2x + 3when x = 4

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### Example:

 Find the equation of the normal to the curve y = x2 - 3x + 5 at the point (1,3) Steps: y = x2 - 3x + 5 1) General normal equation y = mx + c 2) Differentiate to find gradient of tangent m1 y' = 2x - 3 3) Substitute x = 1 into the derivative y' = 2x1 - 3 = -1 4) The result is the gradient of the tangent m1 m1 = -1 5) Find the negative inverse of m1to give the gradient of the normal line m2 m2= 1 Normal equation so far is y = 1x + c 6) Substitute in x=1 & y=3 values in to find c y = 1x + c 3 = 1x1 + c 3 = 1 + c 2 = c 7) Normal equation is y = 1x + 2