How do you graph #y = 10/3x + 5#?

1 Answer
May 18, 2018

Answer:

graph{10x/3 + 5 [-10, 10, -5, 5]}

Explanation:

The equation is in the form, #y=mx+c#
where,
#m rarr# slope of the line
#c rarr# y-intercept

Here, #m = 10/3#
but, slope = tan#theta#
...where, #theta rarr# angle made by the line with positive x axis

#:. tantheta = 10/3#

#:. theta = 73.3# degrees

Now, to find x- intercept, put y = 0
#:. 10/3x+5=0#
#:. 10/3x=-5#
#:. x=-1.5#

This means that at ant the point (-1.5, 0), you have to draw an angle of 73.3 degrees. this line will represent the given equation.

An Alternative method:

Make a table:
Consider any random 3 values of x, and find the corresponding values of y. Plot these points, and th eline will pass through all these points linearly.

E.g.: I consider 3 cases, x=0, =1,=-1
1] For # x=0#, from the given equation, I get #y = 5#, so i plot the point #(0,5)# on my graph.

2] For #x=1#, I will get #y=8.3#. So i plot the point, #(1,8.3)#.

3] For #x=-1#, I will get #y=1.7#. So i plot the point, #(-1,1.7)#.

Once you get these 3 points on your graph, simply draw a line that passes through each one of them.

(You can follow the same 3 steps while assuming any value of y also.)

*Hope this helps :) *