Equation Find 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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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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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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