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Equation of a Tangent line

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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.

 

Class Notes | Blank Notes

Ex15.03,

Nulake IAS p84,85

 

 

 

Find the gradient of the tangent line (m) (differentiate and substitute in the x value)
y =mx + c
 
Substitute the x & y values of the point into the equation to find c

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

 

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Find the equation of the tangent to the curve y = 0.5x2 - 3x +7at the point (4,3)

Mouse over boxes below

The tangent line touches the curve at one point

Refresh the page to redo the mouse over steps

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Class Notes | Blank Notes screen snaps

 

Equation of a Normal Line (extension)

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

Mouse over boxes right

The method is the same as finding the tangent equation apart from the change in gradient required.

NOTE: The gradient of the tangent line and normal line multiply to -1

Refresh the page to redo the mouse over steps

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

 

 

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