Answer:
[tex]y=2x+k[/tex]
Step-by-step explanation:
[tex]y=2x+k[/tex]
Here we are given that the plant growth is fixed by 2 inches for every month.
hence ,
if we assume that the plant had zero height at beginning , when the time was also zero, then in the next month that is 1st month the height of the plant will be 2 inches, similarly in the 2nd month the height will be 4 inches , for the third month the height of the plant will be 6 inches and so on.
So, if we plot them on a graph , taking time in month on x axis and height of the plant on y axis , the pattern of growth as mentioned above will give us few coordinates like,
(0,0) , (1,2) , (2,4) , (3,6) and so on..
hence from here we can determine equation which refers the above situation by using two point form of an equation . The two point form is given as
[tex]\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}[/tex]
where
[tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] are [tex](1,2)[/tex] and [tex](2,4)[/tex] respectively
substituting them in the above formula we get
[tex]\frac{y-2}{x-1}=\frac{4-2}{2-1}[/tex]
[tex]\frac{y-2}{x-1}=2[/tex]
[tex]y-2=2(x-1)[/tex]
[tex]y-2=2x-2[/tex]
[tex]y=2x[/tex]
Hence we get out scenario , in equation as [tex]y=2x[/tex]
Note:
this is the situation when the height of the plant was considered 0 at time 0 . however , if the initial height of the plant was k ( assumption )
The equation would have been like this
[tex]y=2x+k[/tex]
This is because in both the cases the slope of the line would be same , as slope represents the rate of increase of y with respect to x . in this case it would be the increased in the height of the plant with respect to the months Where k represent the value of y at x= 0 , or the height of the plant at time 0 month.
Please refer to the image attached with this.
