大家好,有幾題數想大家幫忙解答
1.find the equations of the tangents to the curve y=x^2 + 4x +1
form an external point (1,2).
2.given a curve C:y=1/2 (x)^3 and a point P(0,-1)
a.determine whether P is a point on the curveC.
b.find the equtaion of the tangent to the curve C from the point P.
3.let f(x)=2x^2 - x + 1. Find the equtaion of the tangent to the curve y = f(x) at the point where y = 2 and f(x) is increasing.
4.let f(x) = x^3 - 3x^2 -9x +4. Find the equation of the tangent to the curve y = f(x) at the point where y = 4 and f(x) is decreasing.
THX!!
中四 M1數一問
2011-04-11 5:19 am
回答 (2)
2011-04-11 6:00 pm
✔ 最佳答案
(1)圖片參考：http://imgcld.yimg.com/8/n/HA00159313/o/701104100129113873424400.jpg
圖片參考：http://imgcld.yimg.com/8/n/HA00159313/o/701104100129113873424401.jpg
(3)
圖片參考：http://imgcld.yimg.com/8/n/HA00159313/o/701104100129113873424412.jpg
(4)
圖片參考：http://imgcld.yimg.com/8/n/HA00159313/o/701104100129113873424413.jpg
希望你睇得明啦 =]
唔知你學左微積分未呢??
參考: 雙照淚痕乾自答
2011-04-11 5:19 pm
1) dy/dx = 2x + 4
Suppose the tangent at (h, k) is the required one, then:
2h + 4 = (k - 2)/(h - 1)
2h2 + 2h - 4 = k - 2
2h2 + 2h - 2 = h2 + 4h + 1
h2 - 2h - 3 = 0
h = -1 or 3
k = -2 or 22
At (-1, -2), slope = 2, so by point-slope form, equation is:
(y + 2)/(x + 1) = 2
y + 2 = 2x + 2
2x - y = 0
At (3, 22), slope = 10, so by point-slope form, equation is:
(y - 22)/(x - 3) = 10
y - 22 = 10x - 30
10x - y - 8 = 0
2a) Sub x = 0, y = 0
Hence P is NOT on C.
b) dy/dx = 3x2/2
Suppose at (h, k), the tangent is the required one, then:
(k + 1)/h = 3h2/2
k + 1 = 3h3/2 = 3k
k = 1/2
h = 1
So equation of the tangent is:
(y - 1/2)/(x - 1) = 3/2
2y - 1 = 3x - 3
3x - 2y - 2 = 0
3) When y = 2:
2x2 - x + 1 = 2
2x2 - x - 1 = 0
x = 1 or -1/2
f'(x) = 4x - 1
So when x = 1, f(x) is increasing and so the tangent is:
(y - 2)/(x - 1) = 3
y - 2 = 3x - 3
3x - y - 1 = 0
4) When y = 4:
x3 - 3x2 - 9x + 4 = 4
x3 - 3x2 - 9x = 0
x = 0
f'(x) = 3x2 - 6x - 9
So when x = 0, f(x) is decreasing and so the tangent is:
(y - 4)/x = -9
y - 4 = -9x
9x + y - 4 = 0
Suppose the tangent at (h, k) is the required one, then:
2h + 4 = (k - 2)/(h - 1)
2h2 + 2h - 4 = k - 2
2h2 + 2h - 2 = h2 + 4h + 1
h2 - 2h - 3 = 0
h = -1 or 3
k = -2 or 22
At (-1, -2), slope = 2, so by point-slope form, equation is:
(y + 2)/(x + 1) = 2
y + 2 = 2x + 2
2x - y = 0
At (3, 22), slope = 10, so by point-slope form, equation is:
(y - 22)/(x - 3) = 10
y - 22 = 10x - 30
10x - y - 8 = 0
2a) Sub x = 0, y = 0
Hence P is NOT on C.
b) dy/dx = 3x2/2
Suppose at (h, k), the tangent is the required one, then:
(k + 1)/h = 3h2/2
k + 1 = 3h3/2 = 3k
k = 1/2
h = 1
So equation of the tangent is:
(y - 1/2)/(x - 1) = 3/2
2y - 1 = 3x - 3
3x - 2y - 2 = 0
3) When y = 2:
2x2 - x + 1 = 2
2x2 - x - 1 = 0
x = 1 or -1/2
f'(x) = 4x - 1
So when x = 1, f(x) is increasing and so the tangent is:
(y - 2)/(x - 1) = 3
y - 2 = 3x - 3
3x - y - 1 = 0
4) When y = 4:
x3 - 3x2 - 9x + 4 = 4
x3 - 3x2 - 9x = 0
x = 0
f'(x) = 3x2 - 6x - 9
So when x = 0, f(x) is decreasing and so the tangent is:
(y - 4)/x = -9
y - 4 = -9x
9x + y - 4 = 0
參考: 原創答案
收錄日期: 2021-04-13 17:55:00
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