If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Linear & nonlinear functions: missing value

Learn to find the missing value in a table to make sure it represents a linear equation. Created by Sal Khan.

Want to join the conversation?

  • male robot donald style avatar for user Weston
    ur voice makes me sleepy sal
    (11 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Zach Higgins
    In the Linear and nonlinear functions exercise, there is a type of question which displays an equation not in linear format and asks if the given equation can be expressed as a linear equation. Here's a link to a screenshot of an example: http://imgur.com/UC1j1su

    I understand that because of the square (in some versions it's a square root) somehow, this precludes linear expression, but I don't understand the given proofs.

    Here they prove it's not on a line by deriving three points, (0, -4)(-1, 1)(3, 5). I see (0, -4) comes from assuming -4 is the y-intercept, but where do the other two coordinates come from?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • duskpin ultimate style avatar for user Bella B
    In this example, isn't it more logical to just take "x = 2" and "y = 3" then multiply it by 4? That way "x = 8" (2 * 4 = 8) and "y = 12" (3 * 4 = 12.)
    This is the method I always use, and I don't understand why it has to be made so much more complicated.
    (4 votes)
    Default Khan Academy avatar avatar for user
  • piceratops tree style avatar for user Paige Kathleen
    Is there any way to tell if something is a linear function without making a graph? If the problem is 'Is y+9x a linear function?', is there a way to do it without a graph?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • piceratops seed style avatar for user narnian
    Wait so it's linear? I'm hopelessly confused. What about the big jump from 5 to 8?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • piceratops seed style avatar for user narnian
    So is the problem linear or non-linear? It's non-linear right? Because it did not make a straight line?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • starky ultimate style avatar for user yanjin
      It did make a straight line so it is linear. If Sal went and added the missing x-axis points 4, 5, 6, and 7, the y-axis points would be 6, 15/2, 9, and 21/2. Even though there was a jump from 3 to 8, if you were to graph it, the line would have a slope of 3/2.
      (3 votes)
  • aqualine ultimate style avatar for user JunkoEnoshima
    I need help with linear bc I in 5th grade
    (2 votes)
    Default Khan Academy avatar avatar for user
  • aqualine seed style avatar for user karenelaine1961
    How go I find the missing coordinate of an equation
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user saridski
    Hi there,

    I am trying to find a way to identify the most aggressive increase in value of a non linear line chart but I am uncertain how to do so. Let's say that I have five moving averages, which form five lines moving in time from 0 and oscillating between negative and positive values. I want to be able to identify the two moving averages, which are increasing and decreasing most aggressively. Due to the fact that the lines are non linear, a linear function will not work but a non-linear may. The most recent data is also of most interest so I am looking for an exponential non-linear function. Does this make sense? Thanks in advance.
    (2 votes)
    Default Khan Academy avatar avatar for user
  • aqualine seed style avatar for user blaynemassengill
    Im confused. How would you solve an Indirect Variation if it was something like Y= X (Exponent of 2) +3?
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

Find the missing value to make the table represent a linear equation. So let's see this table right over here. So when x is equal to 1, y is 3/2. When x is 2, y is equal to 3. So let's see what happened. When x increased by 1, what did y do? Well looks like y increased by 3 and 1/2 is the same thing as 1 and 1/2. So to go from 1 and 1/2 to 3, it increased by 1 and 1/2, or you could say it increased by 3/2. You could say that 3 is the same thing as 6/2, 6/2 minus 3/2 is another 3/2. All right. Now when we go from 2 to 3, we're increasing by 1 again in x. And what are we doing in y? So we're going from 3, which is the same thing as 6/2 to 9/2. So once again, we are increasing by 3/2. So in order for this to be a linear equation or a linear relationship, every time we increase by 1 in the x direction, we need to increase by 3/2. If we increase by 2, we need to increase by 2 times 3/2. So what are we doing over here on this fourth term on the table? Well we're increasing. We're going from 3 to 8 so we are increasing by 5. So if we're increasing x by 5, then we need to increase y by 5 times 3/2 or 15 over 2. So that's the amount that we have to increase y by. If we started at 9/2 and we're going to increase by 15/2, so it's going to be 9/2 plus 15/2, this is how much we increment by. Remember, we increment 3/2 every time x moves 1. This time, x moved 5. So we're incrementing by 15/2, or you could say we're incrementing by 3/2 five times. But this is going to be equal to 9 plus 15 is 24 over 2 which is equal to 12. And so in the box, we could write 12, and we are done.