Formulating quadratic and square-root equations using graphs - Part 2

0:00- [Instructor] We are told, "The table shows some points on a square root function, g(x) which is defined for x>0." All right, "Graph g inverse of x, the inverse of g, then use that graph to write the equation of g(x)." This is interesting. So pause this video, have a go at this before we do this together.

0:20Okay, the first step, they want us to graph the inverse of g. So let's just figure out some of those points. So if this is the inverse, let's make another table right over here. And so if we have x and then the inverse of g(x), so essentially we can just swap these x and g of x values here.

0:40So it'd be 4, 0, 6, 1, 8, 4. Remember, when we think about a function and its inverse, if a function maps from one point to another, then the inverse just maps back from that output point back to that input point. And that's what's happening right here. This goes from 4 to 8, this goes from 8 to 4.

1:04Now we know that if we're taking the inverse of a square root function, what we're going to have is essentially a- Or we're going to have part of a parabola. So let's see if we can graph that. So let's see, we have 4, 0, 6, 1, 8, 4. So we're up operating in the first quadrant, so let me graph that accordingly or set my axes up accordingly.

1:31So that's my x axis, this is my x axis. And then let's see, x goes all the way up to 8 so I need to have some space there. So let's say 1, 2, 3, 4, 5, 6, 7, and 8. And we go up to four on the inverse function. So 1, 2, 3, 4, 1, 2, 3, 4. And now let's graph these points.

2:02We have 4, comma, 0. 4, comma, 0, right over there. Then we have- Let me do some different colors just for kicks. Then we have 6,1, 6,1 is right over there. And then 8,4. 8,4 is right over there. Obviously this hand drawn, it's not gonna be perfect, but then the part that would be the inverse of our quadratic- Or sorry, the inverse of our square root function, which is going to be, you could think of it as the right half of a quadratic of a parabola is right over there.

2:42Now let's figure out, what we're gonna do now is let's figure out the equation that fits this parabola. And our equation is actually not just going to give us that side, it's going to give us the other side too, but we know if when we take the inverse back to go back to the square root function, we're going to have to restrict the- I guess we could say we're gonna be taking the principle root, but let's just figure out the equation of this parabola right over here.

3:10Well, we know that this red point right over here, 4,0 is the vertex. Why do we know that? Well, we know that this is only defined for x>0, so in that square root function, which is gonna look something like that, this point right over here, when you take the inverse becomes the vertex.

3:28We don't have any points to the further left of that. So this is the vertex, we know that the equation of a parabola, or a quadratic in vertex form, is y is equal to a times x, some constant a times x minus the x coordinate of the vertex, which in this case is 4 squared.

3:52That's a different, that's blue, not black, squared, plus the y coordinate of the vertex. And the y coordinate of the vertex here is 0 so we could just write a plus zero or just write nothing at all there. So how do we solve for a? Well, we could use one of the other points.

4:08We know that when x is 6, y is 1. So let's put a 6 here and put a 1 there. And so we get 1 is equal to 6 minus 4 is 2, that squared is 4. So it's equal to 4a, divide both sides by 4, and you get a is equal to 1/4. Now where does this get us? Well, now we get the equation for this entire parabola.

4:33So maybe I'll write it down here. We get y is equal to 1/4 times x minus 4 squared. Now if we wanna figure out the inverse of this, because we wanna figure out the equation for g of x, we can swap the x and the y and then solve for y again. So if we think about the inverse, we'd get x is equal to 1/4 times y minus 4 squared.

5:04I could have, well, I'll stick to this way, I could have used the function notation, but I find it a little bit easier to work with my xs and ys for now. And then let's solve for y. So we can multiply both sides by 4. Let's do that. And we're going to get 4x is equal to, that and that cancels, is equal to y minus 4 squared.

5:27Now I can take the principle root of both sides and I am going to get, on the left hand side, if I take the square root of this, the square root of 4x is the same thing as 2 times the square root of x, I'm just bringing the four out, 4 is 2 squared. And so this is going to be equal to y minus 4.

5:48And now I can just add 4 to both sides and I get 2 times the square root of x is equal to, whoops- Plus 4 is equal to y, or I could write, 'cause y is equal to g(x) in this case, g(x) is equal to 2 times the square root of x plus 4. And we're done. And if we want to make sure that we didn't do something strange with the principle root or we're taking the wrong half of the parabola somehow, we could verify that these points indeed satisfy.

6:190,4, x is 0, then we're just going to get a 4. G of g 0 is 4. 1,6, so 2 times 1 plus 4 is 6. 4,8, 2 times the square root of 4 is 4 plus 4 is indeed 8, so we're looking good.