Please help with this question. Thanks.?
2016-05-31 11:55 am
Find an equation of the tangent line to the hyperbola y = 3/x at the point (3,1)
回答 (5)
2016-05-31 11:58 am
✔ 最佳答案
y = 3/xdy/dx = -3/x^2
dy/dx(at x = 3) = -3/3^2 = -1/3
The tangent line is
y - 1 = -1/3(x - 3)
y - 1 = -x/3 + 1
y = -x/3 + 2
2016-05-31 12:17 pm
y=3/x
y' = -3/x^2
when x=3, y'=-3/3^2 = -3/9 = -1/3
Equation of tangent at (3,1):
y-1 = (-1/3)(x-3)
y=(-1/3) x + 1+1
y=(-1/3) x +2
y' = -3/x^2
when x=3, y'=-3/3^2 = -3/9 = -1/3
Equation of tangent at (3,1):
y-1 = (-1/3)(x-3)
y=(-1/3) x + 1+1
y=(-1/3) x +2
2016-05-31 12:01 pm
y = 3/x
y = -3/x²
Slope of the tangent at the point (3, 1)
= y | (x = 3)
= -3/(3)²
= -1/3
The equation of the tangent at the point (3, 1) in point-slope form :
y - 1 = (-1/3)(x - 3)
3(y - 1) = -(x - 3)
3y - 3 = -x + 3
x + 3y - 6 = 0
y = -3/x²
Slope of the tangent at the point (3, 1)
= y | (x = 3)
= -3/(3)²
= -1/3
The equation of the tangent at the point (3, 1) in point-slope form :
y - 1 = (-1/3)(x - 3)
3(y - 1) = -(x - 3)
3y - 3 = -x + 3
x + 3y - 6 = 0
2016-05-31 12:00 pm
Well,
first we verify that point A(3, 1) IS on the graph as it verifies the equation
the slope of the tangent is the derivative :
dy/dx = -3/x^2
therefore the slope is
m = dy/dx(3) = -3/9 = -1/3
therefore the equation is :
y - 1 = (-1/3)(x - 3)
-3y + 3 = x - 3
x + 3y = 6 <--- answer
hope it' ll help !!
first we verify that point A(3, 1) IS on the graph as it verifies the equation
the slope of the tangent is the derivative :
dy/dx = -3/x^2
therefore the slope is
m = dy/dx(3) = -3/9 = -1/3
therefore the equation is :
y - 1 = (-1/3)(x - 3)
-3y + 3 = x - 3
x + 3y = 6 <--- answer
hope it' ll help !!
2016-05-31 11:58 am
y = 3/x
y' = -3/x^2
y' = -3/(3^2) = -3/9 = -1/3
y - 1 = (-1/3) * (x - 3)
y - 1 = (-1/3) * x + 1
y = (-1/3) * x + 2
y' = -3/x^2
y' = -3/(3^2) = -3/9 = -1/3
y - 1 = (-1/3) * (x - 3)
y - 1 = (-1/3) * x + 1
y = (-1/3) * x + 2
收錄日期: 2021-04-18 14:55:08
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160531035513AAY8QSN