keeping in mind that parallel lines have the same exact slope, hmmmm what's the slope of the line above anyway?

so we're really looking for the equation of a line whose slope is 1/3 and runs through (18,2)

First write it in vertex form :-
y= a(x - 2)^2 + 3 where a is some constant.
We can find the value of a by substituting the point (0.0) into the equation:-
0 = a((-2)^2 + 3
4a = -3
a = -3/4
so our equation becomes y = (-3/4)(x - 2)^2 + 3
Answer:
you plot the coordinates you have been given
Step-by-step explanation:
A or b depending on wich way your moving