How to use Newton's method to find where 2 curves intersect

1) the first set of 2 curves is y=x^2 and y=x+2
2) second set of curves is y=e^-x and y=lnx correct to 3 decimal places
i just really need to know how to do this as it starts off with finding the gradient or something using : y-y0=m(x-x0) and i dont know what to do
please show working

1) Equation is x^2-x-2=0 and f(x)=x^2-x-2
You want f '(x)=2x-1.
You require an approximation to the required value of x; say x=1.5 (here the exact values
are x=2 and x=-1, so using Newton is a sledgehammer to crack a nut)
Referring to this value as x1, the next approximation is given by
x2=x1 - f(x1)/f '(x1) =1.5 - [1.5^2-1.5-2]/[3-1]=2.125 then repeat to find
x3=2.0048, x4=2.00000768 and it clearly homes in on 2
Having found x find y from one of the equations.

2) For e^-x=lnx, use f(x) = lnx - e^-x, f '(x)=1/x+e^-x
You need a first approx which you can obtain by sketching y=e^-x and y=lnx
x=1.5 should do.
x2=1.5 - [ln1.5 - e^(-1.5)]/[1/1.5+e^(-1.5)]=1.2905
x3=1.30971, x4=1.309799583,x5=...
I use a calculator with Ans function which enables you to calculate these
very easily

They intersect... so they are equal in y-value

x² = x + 2

So
x² - x - 2 = 0 .......... (simple quadratic)

This is set equal to zero, as you can see... so this is the root, or zero, of the function
y = x² - x - 2

Simple enough? Apply Newtons method.

ahh numerical methods, skip!