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### Course: Functions 229-236>Unit 1

Lesson 3: Intercepts

# Worked example: intercepts from an equation

Let's find the x- and y-intercepts of the equation 2y + 1/3x = 12, where a line crosses the x-axis (the x-intercept) and the y-axis (the y-intercept) on a graph. To find the x-intercept, we make y equal to 0 and solve the equation for x. To find the y-intercept, we make x equal to 0 and solve for y. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• I know in the equation y=mx+b, b is the y-intercept. Is there a x-intercept in another slope equation? Thanks!
• Slope intercept form is y=mx+b. Really, it is more like "Rise/Run Slope of a line from the y-intercept form". If you want to know the x-intercept form you can solve the equation for x.

y=2x+4 would become x=1/2y-2. Now plug in zero for y and you will have what x is equal to (-2). The reason we use y=mx+b is because the idea of rise/run no longer works correctly in this version of slope x-intercept. You have run/rise instead of rise/run. This is unwanted confusion as now you have

y=mx+b "Rise/Run Slope of a line from the y-intercept form"
and
x=my+b "Run/Rise Slope of a line from the x-intercept form"

I hope this helped you find the x-intercept and also why we only use y-intercept slope formulas.
• How do you find intercepts when the formula is more complex as in y=x*sqrt(16-x^2)
• The general rule of thumb for finding intercepts is to plug 0 in for the other number. To finish the equation you provided, you would need to use the zero-product property to set each piece equal to zero and solve..
• Need help, what do the "|" mean in an equation? Example, y=|x-2|
Thanks
• those are called absolute value bars, no matter what is inside the absolute value bars, it comes out as positive. So, for example |4| = 4 and |-4| = 4.
• So, if you wanted to do the same (figure out the x and y intercept) for any other equation, should I only use 0 as a substitute to figure them out? Would it be the same if you used other numbers as substitutes?
• Katie, yes, exactly. If the equation is in standard form:ax+by=c then you just set x equal to 0 to find the y intercept and set y equal to 0 to find the x intercept. It makes sense, right. The y intercept is when the line crosses the y axis, so x will always be 0, so that is why we set x equal to 0. To find where the line crosses the y axis.

The same logic applies when finding the x intercept, set y equal to 0 to find where the line crosses the x axis, because y will always be zero at the x intercept.
Ex: (0,8) y intercept, notice x is 0
(2,0) x intercept notice y is 0
• Is there a rule governing linear coordinates that pass the exact centre of the graph?
For instance, where Xintercept=Yintercept=0...

Since you can't draw a linear graph from using just one point, what can we do with such information?
I understand a lot of formulas will meet with the centre, and the lines can angle 360degrees,
through all quadrants of the graph when it does.
But is it some type of typical linear graph? Or is there special methods involved?
What can we know about situations that don't work while originally meeting the demands for the X/Yintercept method? (crossing both x0 and y0, being a linear equation etc.)
Perhaps it is a singular exception?

Sorry for this weakly translated and confusing question.
I'd like to hear anything people got to say about this.
• When linear coordinates pass the exact centre of the graph, i.e. x-int=y=int=0; you cannot draw a line just based off this information due to the fact that it is quintessential that you have or know a minimum of 2 points in order to graph the line. Factors of a line which will help in this scenario will include; the gradient, the equation of the line and even just simple another point. Hope this helps!
• How do answer questions about slope in word problems? That is my problem on math test with any unit.
• (1 vote)
• Can X and Y be negative
• Simple answer, Yes
• this doesn't relate to this vid at all but dividing anything by zero is so random like 1/0.0001=10000 and 1/0.000000=1000000. so does that mean 1/0 is infinity?
• What you're doing is doing `1/x` when `x` is approaching `0`. If `1/0` were to have a value, this is a totally valid way to go about it, BUT we have to see if the result is the same whether approaching from the positive side or the negative side.
Problem is you're only approaching it from the positive side and haven't checked the negative side yet!

Ok! If we make `x` approach `0` from the negative side in the same manner, we get `1/-0.1 = -10, 1/-0.01 = -100, 1/-0.0001 = -10000, ...`. As you can see, the value of `1/x` is getting larger and larger on the negative side, so it's approaching negative infinity.
Hold on.
- When we evaluated `1/x` as `x -> 0+`, it approached positive infinity.
- When we evaluated `1/x` as `x -> 0-`, it approached negative infinity.

So, does `1/0 = infinity`?