Find the equation of the tangent to the curve y = x^3 - 7x^2 + 14x - 8 at the point where x=1.
3 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
The slope of the tangent to any point on a curve is given by the value of the first derivative of the curve at the point. For y=x^3 - 7x^2 + 14x – 8
y’ = 3x^2 – 14x + 14
At the point where x = 1.
y’ = 3*1 – 14*1 + 14
=> y’ = 3
Now the value of y at the point where x = 1 is y = 1^3 – 7*1^2 + 14*1 – 8 = 1 – 7 + 14 – 8 = 0.
The equation of the line with a slope equal to 3, passing through (1, 0) is given as
(y – 0)/(x – 1) = 3
=> y/ (x – 1) = 3
=> 3x – 3 = y
=> 3x – y – 3 = 0
The required equation of the line is 3x – y – 3 = 0.
giorgiana1976 | College Teacher | (Level 3) Valedictorian
We'll determine the y coordinate of the tangency point, that is:
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
neela | High School Teacher | (Level 3) Valedictorian
To find the tangent at x=1 to the curve y = x^3 - 7x^2 + 14x - 8.
The tangent to a curve y = f(x) at x1 is y-y1 = f'(x1)*(x-x1), where y1 = f(x1).
y = f(x) = y = x^3 - 7x^2 + 14x - 8.
dy/dx = f'(x) = y = (x^3 - 7x^2 + 14x - 8)'
f'(x) = 3x^2-14x+14.
f'(x1) = f'(1) = (3*1^2-14*1+14) = 3.
y1 = f(1) = 1^3-7*1^2+14*1-8 = 0
Therefore the tangent at (x1,y1) = (1,0) is given by:
y-0 = 3(x-1).
y -3x +3 = 0.