Question

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

Answer 1

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 :) *