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

Graphing the inverse of a linear function

Sal is given a line segment on the coordinate plane, and he graphs the inverse of the function represented by that segment.

Want to join the conversation?

  • male robot johnny style avatar for user ϟ 2-XL ϟ
    Are all linear functions invertible around y=x?
    (22 votes)
    Default Khan Academy avatar avatar for user
  • male robot hal style avatar for user RandomDad
    Sal said in his first video in inverse function that he will explain why he constrained the domain of function. Where is the video that explain the reason ?
    (11 votes)
    Default Khan Academy avatar avatar for user
  • male robot johnny style avatar for user kallen.watson
    Will the line Y=X stay the same or would something strange happen?
    (7 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user loananhdp
    Can someone please explain to me the concept of the horizontal line test? I'm doing the domain and range with inverse functions exercise.
    (3 votes)
    Default Khan Academy avatar avatar for user
    • duskpin ultimate style avatar for user jewelinelarson
      The horizontal line test is used for figuring out whether or not the function is an inverse function. Picture a upwards parabola that has its vertex at (3,0). Then picture a horizontal line at (0,2). The line will touch the parabola at two points. This is how you it's not an inverse function.
      (5 votes)
  • blobby green style avatar for user Tatum Raye Lynn
    So i have never understood how to get Domain and Range of anything or where it comes from or why. I've tried again again to understand.... Maybe you could help me ?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • orange juice squid orange style avatar for user Edward Moore
      TRL, I think of domain as the possible values of x that will result in a point on a curve; and range are the possible values of y that could result in a point on a curve. Think of a parabola... any x can be substituted into the equation to get a y on the curve, so the domain is all real numbers. But our parabola has a minimum or maximum, so there's a point where the y-values stop being an option. That's a restriction on the range, that y can only be greater than or equal to the minimum, or less than or equal to the maximum, depending on the way it's facing. In this inverse video, the trick is to remember that the range of a function affects the domain of the inverse, and the domain of a function affects the range of the inverse :-)
      (2 votes)
  • blobby green style avatar for user MakenzieMcLane
    how do you solve the inverse function of y=x^2 +2
    (1 vote)
    Default Khan Academy avatar avatar for user
  • sneak peak green style avatar for user Shinichi Kudo
    If the graphs of a function and its inverse function intersect, would the two graphs intersect on the line y=x? What is the possibility for the two graphs to intersect on other lines?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • mr pink green style avatar for user David Severin
      Since the idea that you switch x and y, it makes sense that the primary place that this happens is along the y=x (or if you switch places, x=y) line. They could possibly intersect on an infinite number of other lines because any point has an infinite number of lines that go through it, but it has to be on the y=x line. Say the point of intersection is (2,2) this is on the y=x line, but is it also on the y = 4x - 6, y = 3x-4, etc. lines, but these are just arbitrary lines that just happen to contain the point (2,2) but are otherwise totally unrelated to the inverse function. Because they reflect across the y=x line, one rule of reflection is that if a point is on the line of reflection, then it maps back to itself. That would be true of my other examples, but none of the other points on the function and its inverse would reflect across these line.
      (1 vote)
  • piceratops tree style avatar for user ŇØŦ€ŞŁΔ¥€Ř
    Implicitly defined curves do have "inverses" but I don't understand why they may not be the graph of a function. For example, if I plot the inverse relation of f(x) = x^3-4x, how can you say it is not the graph of a function? It would really help me out if you could answer this question. Thank you : )
    (2 votes)
    Default Khan Academy avatar avatar for user
    • mr pink green style avatar for user David Severin
      So the function you present has an inverse that is not a function. If you factor this out to x(x^2-4) or x(x-2)(x+2) you have three x intercepts at (0,0) (2,0) and (-2,0). If you switched x and y, your inverse function would have 3 y intercejpts at (0,0) (0,2) and (0,-2) which does not pass the vertical line test and has one x value go to 3 different y values, and is thus not a function. Some cubic functions such as the parent function (y=x^3) or with just a cube and constant (y=x^3 - 27) would have inverses that are functions.
      By starting out with a function, you know it passes the vertical line test by definition. However, for its inverse to be a function, it also has to pass the horizontal line test to insure the one-to-one correspondence of x and y values.
      (2 votes)
  • orange juice squid orange style avatar for user Genevieve Krause
    If h^-1(x) is the inverse of h(x) then are h^-1(y) and h(x) the same?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • duskpin sapling style avatar for user mthompson2019
    How can you reflect a line across y=x without graphing each point? Is there any way to do it in a graphing calculator?
    (2 votes)
    Default Khan Academy avatar avatar for user

Video transcript

This right over here is our understanding inverses of functions exercise on Khan Academy. It's a good exercise to make sure you understand inverses of functions. It's an interactive one where we can move this line around and it tells us 'the graph of h(x) is the green', so that's this dotted green line, 'the dashed line segment shown below'. So that's this. 'Drag the endpoints of the segment below to graph h inverse of (x). There's a couple of ways to tackle it. Perhaps the simplest one is we say, okay, look, h(x), what does h(x) map from and to? So h(x), this point shows that h(x), if you input -8 into h of (x), h of -8 is 1, so it's mapping from -8 to 1. Well, the inverse of that, then, should map from 1 to -8. So let's put that point on the graph, and let's go on the other end. On the other end of h of x, we see that when you input 3 into h of x, when x is equal to 3, h of x is equal to -4. So this point shows us that it's mapping from 3 to -4. So the inverse of that would map from -4 to 3. If you input -4 it should output 3. Since we took the two end points of this line and found the inverse mapping of it, what I have just done here is that I have graphed the inverse. Another way to think about the inverse is if you were to draw the line y = x, these things should be reflections around the line y =x because one way to think about it is, you're swapping the xs for the ys. If you were to draw the line y = x, if you flipped it around, the line y = x, the green line, you would actually get the old line. This would flip over there and this would flip over there. But either way, we're done. We have graphed h inverse of x.