- 1 ) we get 3 * 3 + 2 * ( - 1 )= k ..k = 9 - 2 =7..Hence the required equation is3x + 2y = 7 .______________________________________….......
f'(x) = 4x - 8 ← this is the slope of the tangent line to the curve @ x
f'(x) = 12 - 8 = 4 ← this is the slope of the tangent line to the curve @ x = 3
The equation of the tangent line becomes: y = 4x + b
You know that the tangent line passes through (3 ; - 6), so these coordinates must verify the equation of the tangent line.
y = 4x + b
b = y - 4x → you substitute x and y by the coordinates of the point (3 ; - 6)
b = - 6 - (4 * 3) = - 6 - 12 = - 18
→ The equation of the tangent line to the curve is: y = 4x - 18
1). Any st line parallel to the line 3x + 2y = 4 will be of the form
.
3x + 2y = k
.
When this passes through the point ( 3 , - 1 ) we get 3 * 3 + 2 * ( - 1 ) = k .
.
k = 9 - 2 = 7.
.
Hence the required equation is 3x + 2y = 7 .
______________________________________…
.
2). 4cos 3x + 7 = 6
.
4 cos 3x = 6 - 7 = - 1
.
cos 3x = - 1/4
.
3x = cos^-1 ( - 1/4 ) = pi +/- cos^-1 (1/4 )
.
x = 1/3 [ pi + / - cos^-1(1/4)
--------------------------------------…
General solution:
............. 3x = (2n+1)pi + / - cos^-1(1/4)
..
................... x = 1/3 [ ( 2n+1 ) pi + / - cos^-1(1/4) ].
.
______________________________________…
.
3). y = 2x ( x - 4 )
.
dy / dx = 2 [ x - 4 + x ] = 2 [ 2x - 4 ] = 4 [ x - 2 ]
.
At x = 3 we have dy / dx = 4 ( 3 - 2 ) = 4 * 1 = 4
.
Hence slope of the tangent at x = 3 will be = 4 .
.
Also at x = 3,......... y = 2 * 3 ( 3 - 4 ) = 6 * - 1 = - 6
.
Hence the equation of the tangent to the curve at x=3 , where y = - 6 and slope of the tangent = 4 will be
.
y + 6 = 4 ( x - 3 )
.
That is y + 6 = 4 x - 12
.
That is 4 x - y - 18 = 0 is the required equation .
1.
3x + 2y = 4
2y = -3x + 4
y = -3x/2 + 2
For the line to be parallel, it must have the same slope (coefficient of 'x' which is '3/2)/
