The line y = px + 1 passes through (0,1) so its y intercept is 1.

p can be positive, 0 or negative.

If it is parallel to AB, it will only cut the function at its downward sloping portion ,so only cuts it at 1 point.

And p would be equal to the gradient of AB, which is 2. So p = 2 here.

If it is steeper than that (p > 2), it also only cuts at 1 point. Either the downward sloping or upward sloping portion, but not both.

Likewise, if it is parallel to CD, it will only cut the function at its upward sloping portion. so only cuts it at 1 point.
p would be equal to the gradient of CD, which is -2. So p = -2 here.

If it is steeper than -2 (p< - 2), it also only cuts at 1 point, at only the upward sloping portion

However, if -2 < p < 2, you will see that the line cuts the function at 2 points.

So -2 < p < 2

i need part ii not part i lol

I know. I'm not done explaining. The function doesn't pass through (0,1) anyway so it's evident I wasn't referring to i)

Done.

I write up later

I will recommend you to use the desmos graphing website to visualise this.

make the graph of y = 4 - |2x - 1| and y = px + 1.

There will be a slider function for p, where you can adjust its value and see how the line changes.

https://www.desmos.com/calculator