If P=40n+300, then when n=0 p=300 so the p intercept is indeed 300 and the slope (gradient) is 40. Looks like your book has a mistake?
i'm working pre-algebra on my own so i can take pre-calc and then calc. I am going to post whenever i run into problems. heres one from chap 4, Sec 4.5.
The book graphs the p intercept as 0 but i get 300. heres the problem.
15. Graph each formula for the given values of the independent variable.
P = 40n + 300, 0 ≤ n ≤ 200
i get the ordered pair as ( n, P ),
the slope as 40n,
and the P int as ( 0, 300)
this concludes me to graph the Y or P int at 300, but the answer at the back of the book graphs at 0 w/ the slope being 40/1. Can someone explain why the P int is at 0 and not at 300.