A. Yes you are correct that the gradient at any point is 3/(3x-1). However at point P it would be 3/(3*2-1)=2/5 b. The gradient of the normal would therefore be -5/2 We can use the general formula of an equation to get y-ln(5)=-5/2 (x-2) Now multiply both sides by 2 to get: 2y-2ln(5)=-5x+10 Now when it crosses the x axis we know that y=0 therefore: 5x=10+2ln(5) Therefore: x=2+2/5 ln(5) when y=0 You could find an estimate of this number to be 2.64 (3sf) but this might not be sufficient
