f(x) = 2x² ← this is the parabola

f(3) = 2 * 9 = 18 → the parabola passes through A (3 ; 18), so its tangent line too

f'(x) = 4x ← this is the derivative

…and the derivative is the slope of the tangent line to the curve at x

f'(3) = 4 * 3 = 12 ← this is the slope of the tangent line to the curve at x = 3

Equation of the tangent line

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

You know that the slope of the tangent line is 12.

The equation of the tangent line becomes: y = 12x + b

The tangent line passes through A (3 ; 18), so these coordinates must verify the equation of the tangent line.

y = 12x + b

b = y – 12x → you substitute x and y by the coordinates of the point A (3 ; 18)

b = 18 – 36 = – 18

→ The equation of the tangent line is: y = 12x – 18

Intersection between the tangent line to the curve and the x-axis: → when y = 0

y = 12x – 18 → when y = 0

12x – 18 = 0

12x = 18

x = 3/2

→ Point B (3/2 ; 0)

Intersection between the vertical line passes through the point A and the x-axis: → when x = 3

→ Point C (3 ; 0)

The equation of the vertical line is: x = 3

Area of the region bounded by the parabola y = 2x², the tangent line to this parabola at (3 ; 18), and the x-axis.

= (area of the region bounded by the parabola y = 2x² and the x-axis) – (area of the triangle ABC)

= [∫ (from 0 to 3) of the parabola] – [(xC – xB).(yA – yC)/2]

= [∫ (from 0 to 3) 2x².dx] – [(xC – xB).(yA – yC)/2]

= { [(2/3).x³] from 0 to 3 } – { [3 – (3/2)].(18 – 0)/2 }

= [(2/3) * 3³] – { [(6/2) – (3/2)] * 9 }

= [(2/3) * 27] – { [(3/2) * 9 }

= 18 – (27/2)

= (36/2) – (27/2)

= 9/2 square unit