Math Question-calaulus

Find an equation of the line that is tangent to the graph of f and parallel to the given line.
1.)
Function: f(x)=-(1/4)x^2
Line: x+y=0

y = -1/4 x^2
dy/dx = 2(-1/4)x
dy/dx = -1/2x
That is slope m = -1/2x
The line y = -x
Since the tangent is parallel to y = -x
The tangent should have a slope -1
-1 = -1/2 x
x = 2
If x =2, y = -1/4 x^2
y = -1/4(2)^2 = -1
The point is (2, -1)
The line has slope -1 and passing through (2, -1)
y – (-1) = -1(x -2)
y +1 = -x +2
y = -x + 1
The tangent line is y = -x +1


2.)
Function: f(x)=-(1/2)x^3
Line: 6x+y+4=0
The line is y = -6x -4
F(x) = y = -(1/2)x^3
dy/dx = -3/2x^2
This is also the slope of the line
Slope of the line y = -6x -4 is -6
-3/2x^2 = -6
x^2 =4
x = 2 or x = -2
If x = 2, y = -(1/2)(2)^3 = -4
If x = -2, y = -(1/2)(-2)^3 = +4
The two points are (2, -4) and (-2, 4)
They have slope -6
The line has slope and passing through (2, -4)
The other line has slope and passing through (-2, 4)
y – (-4) = -6(x -2)
y + 4 = -6x + 12
y = -6x +8

y -4 = -6(x –(-2))
y = -6x -12
y = -6x - 8



2010-10-14 08:19:59 補充：
correction: missing -6 in question 2

The line has slope -6 and passing through (2, -4)
The other line has slope -6 and passing through (-2, 4)