a curve C : y=x^3 +ax^2 + bx + c , where a,b and c are constants , touches the x-axis at x=1 and intersects the y-axis at (0,1).
(a) Find the values of a,b and c.
(b) Find the turning point(s) of curve C.
math m2!!
2010-08-20 7:19 am
回答 (2)
2010-08-20 7:36 am
2010-08-20 7:42 am
a)
first sub x = 0, y = 1 into C
you can find c = 1.
as C touches at x- axis at x = 1
sub x = 1,y = 0
then 0 = 1^3 + a(1) +b(1) + 1
a + b = -2 --------(1)
as C touches x-axis at x=1, slope at that point is 0
diff C, dy/dx = 3x^2 +2ax +b
0 = 3 + 2a + b
2a + b = -3 ----------(2)
by solving (1),(2)
a = -1
b = -1
b)
C: y = x^3 -x^2 - x +1
diff C dy/dx = 3x^2 -2x -1
when dy/dx = 0(since slope of turning point is 0)
0 = 3x^2 -2x -1
x = 1 or x =-1/3
when x = 1, y = 0
when x = -1/3, y = 1.185185185.....
since it is degree 3, the C had two turning point
and those two points are the turning point.
but you should check by yourself by sub x<-1/3, 1 > x >-1/3 , x>1 to check the trend of the curve and confirm the ans.
first sub x = 0, y = 1 into C
you can find c = 1.
as C touches at x- axis at x = 1
sub x = 1,y = 0
then 0 = 1^3 + a(1) +b(1) + 1
a + b = -2 --------(1)
as C touches x-axis at x=1, slope at that point is 0
diff C, dy/dx = 3x^2 +2ax +b
0 = 3 + 2a + b
2a + b = -3 ----------(2)
by solving (1),(2)
a = -1
b = -1
b)
C: y = x^3 -x^2 - x +1
diff C dy/dx = 3x^2 -2x -1
when dy/dx = 0(since slope of turning point is 0)
0 = 3x^2 -2x -1
x = 1 or x =-1/3
when x = 1, y = 0
when x = -1/3, y = 1.185185185.....
since it is degree 3, the C had two turning point
and those two points are the turning point.
but you should check by yourself by sub x<-1/3, 1 > x >-1/3 , x>1 to check the trend of the curve and confirm the ans.
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