Help with Higher Maths Questions

1) Find the equation of the line passing through the point (3,-1) which is parallel to the line with equation 3x+2y=4.

2) Solve the equation 4cos3x+7=6

3) Find the equation of the tangent to the curve y=2x(x-4) at the point where x=3

1) slope of lines parallel have the same slope.
slope = -2/3
y = -2x/3 + c
putting (3, -1) we get
-1 = -2 + c
c = 1
equation of the line
y = -2x/3 + 1

2) 4cos3x+7=6
cos3x = -1/4
3x = 135 and 225 degree

x = 45 and 75 degree

3) slope of tangent is dy/dx at x = 3
dy/dx = 4x - 8 = 4

value of y at x = 3
y = 3*2(3 - 4) = -6

equation of the line
y = 4x + c
-6 = 12 + c

c = -18
equation of tangent
y = 4x - 18

The typical equation of a line is: y = mx + b → where m: slope and where b: y-intercept


1) Equation of the line passing through the point (3 ; - 1) which is parallel to the line: 3x + 2y = 4

(ℓ1) : 3x + 2y = 4

(ℓ1) : 2y = - 3x + 4

(ℓ1) : y = - (3/2)x + 2 ← the slope is - (3/2)


Two lines are parallel if they have the same slope. So the slope of the line (ℓ2) is - (3/2).

The equation of the line (ℓ2) becomes: y = - (3/2)x + b

You know that the line (ℓ2) passes through (3 ; - 1), so these coordinates must verify the equation of the line (ℓ2).

y = - (3/2)x + b

b = y + (3/2)x → you substitute x and y by the coordinates of the point (3 ; - 1)

b = - 1 + [(3/2) * 3] = - (2/2) + (9/2) = 7/2

→ The equation of the line (ℓ2) is: y = - (3/2)x + (7/2)


2) Solve the equation

4cos(3x) + 7 = 6

4cos(3x) = - 1

cos(3x) = - 1/4

3x ≈ 104,4775°

x ≈ 34,8258°


3) Equation of the tangent line to the curve: y = 2x(x - 4) @ the point x = 3

f(x) = 2x(x - 4) ← this is the curve

f(x) = 2x² - 8x

f(3) = 18 - 24 = - 6 → the curve passes through the point (3 ; - 6)