What are the intercepts for #y = 6x + 8#?

1 Answer

We find this out by setting either x or y to zero and solving the equation

Explanation:

The x-intercept is the point on a line where it crosses the x (horizontal) axis. That is, y = 0 at that point

graph{y=6x+8 [-15.48, 6.72, -0.9, 10.2]}

So, if we set y = 0, the equation becomes

#0 = 6x + 8#

Solving for x by subtracting 8 from both sides of the equation:

#-8 = 6x#

and divide both sides by 6

#- 8/6 = x#

#x = -1.333... -># this is the #x#-intercept

We can do the same thing for the y- intercept, which is the point where the line crosses the y (vertical axis), and x = 0

#y = 6(0) + 8#

#y = 0 + 8#

#y = 8 ->.# this is the #y#-intercept.

We can also take a shortcut... the equation of a line is:

#y = m(x) + b#

Where #m# is the slope of the line, and #b# is the #y#-intercept. So, in:

#y = 6x + 8#

The #y#-intercept is #8#. Note that this only works when you have the equation in the form #y = m(x) + b#

Check the graph. Do these answers look about right? Does the line cross the x-axis at about #-1.33#? Does it cross the y-axis at around #8#?