Main content
8th grade
Course: 8th grade > Unit 3
Lesson 13: Linear and nonlinear functions- Recognizing linear functions
- Linear & nonlinear functions: table
- Linear & nonlinear functions: word problem
- Linear & nonlinear functions: missing value
- Linear & nonlinear functions
- Interpreting a graph example
- Interpreting graphs of functions
- Linear equations and functions: FAQ
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice
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?
ur voice makes me sleepy sal(16 votes)
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?(5 votes)- So it's not a liner if it doesn't show a line(5 votes)
So is the problem linear or non-linear? It's non-linear right? Because it did not make a straight line?(4 votes)
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.(5 votes)
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.(6 votes)
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?(6 votes)
I need help with linear bc I in 5th grade(3 votes)- Wait so it's linear? I'm hopelessly confused. What about the big jump from 5 to 8?(2 votes)
- You'd divide the change in y with the change in x to find the constant.(1 vote)
Wouldn't this table be non-linear since since the change in x and y wasn't constant? For example the change in x being +1, +1, and then +5
Or is it constant since the change in y also increased by 5 at the same time as x?(2 votes)- You can't determine whether the points give a linear or non-linear graph just from the x values. You'll need to see how the y values change in response to a change in x values. So, see how as x increases by 1, y increases by 3/2. So, if x increases by 5, y must increase by 15/2 for the points to represent a linear equation (See that the whole purpose of the problem is to make the points part of a line)(1 vote)
- 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) - Ppl who is looking at this message then good luck and love to ya. <3(2 votes)
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.