EquationFind the equation of the tangent to the curve y = x^3 - 7x^2 + 14x - 8 at the point where x=1

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find the tangent to the curve y = x^3 - 7x^2 + 14x - 8 at the point where x=1

At x = 1, y = 1 - 7 + 14 - 8 = 0.

The slope of the tangent to the curve at x = 1 is the value of y' at x = 1

y' = 3x^2 - 14x + 14

At x = 1 it is 3 - 14 + 14 = 3

The equation of the tangent is (y - 0)/(x -1) = 3

=> y = 3x - 3

=> 3x - y - 3 = 0

The equation of the tangent is 3x - y - 3 = 0

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll compute y coordinate of the tangency point:

y = 1^3 - 7*1^2 + 14*1 - 8

y = 1 - 7 + 14 - 8

y = 0

So, the tangency point has the coordinates (1,0).

Now, the expression of the first derivative represents the tangent line to the given curve.

y' = x^3 - 7x^2 + 14x - 8

y' = 3x^2 - 14x + 14

For x = 1 => y' = 3 - 14 + 14

y' = 3

The slope of the tangent line is m = 3.

The equation of the tangent line, whose slope is m = 3 and the point of tangency is (1,0), is:

y - 0 = m(x - 1)

y = 3(x - 1)

y = 3x - 3

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