Which part?

y = x + ln(3x + 4)

Point A is on the x-axis, so its y-coordinate is 0.


When y = 0, 0 = x + ln(3x + 4)

ln(3x + 4) = -x

3x + 4 = e^(-x)

3x + 4 = 1/(e^x)

3x e^x + 4e^x = 1


Now this equation can't really be solved manually, so use GC.

Using GC,

x ≈ -0.677208
= -0.677 (3s.f)


You may also choose to solve 0 = x + ln(3x + 4) directly using GC instead of manipulating the terms to get 3x e^x + 4e^x = 1

ii)

y = x + ln(3x + 4)

Now ln(3x + 4) is defined when 3x + 4 > 0 , x ∈ R

The term inside the ln brackets cannot be 0 or negative.


So 3x + 4 > 0
3x > -4
x > -4/3

So the curve will never intersect the line x = -4/3 since all values of x (all the x-coordinates of the points on the curve) have be bigger than -4/3 for the ln(3x + 4) to be defined.

There are no points on the curve where the x ≤ -4/3. The curve will get closer and closer to the line x = -4/3 but never meet or touch it.

Equation of vertical asymptote is x = -4/3
or x = -1⅓

Thank you so much!!