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Polynomial Tangent and Normal Lines
The equation of the tangent to the curve y = 6 x3 – 2 x2 + x – 4 at the point with
x coordinate -2.1 is y = x +

tickcross


Worked Solution

To find the equation of the tangent at ( –2.1, y) we needDifferentiating the equation of the curve gives a formula for the curve's gradient at this point:
dy
---
dx
 = 18 x2 – 4 x + 1

So the gradient of the curve when x = –2.1 is 88.78.
The equation of the tangent is therefore y = 88.78 x + c

To find the value of c, we need a point on the tangent but it goes through ( –2.1, y) on the curve.
The value of y on the curve with x = –2.1 is –70.486.
Putting this into the equation of the tangent gives c = –70.486 + (–88.78)×(–2.1) = 115.952.
The equation of the tangent to the curve at x = –2.1 is therefore y = 88.78 x + 115.952


© MEI Produced by Dr Ron Knott, 23 March 2004
tom . button [AT] mei . org . uk 		Test reference: PolyTanNorms.html?ref=7942