00:02
In this problem, we are given the graph of a function f.
00:07
We are asked to draw the graphs of function which are obtained from f.
00:15
In part a, we're given that function y is equal to f of x minus 2.
00:21
I'll be drawing this function in red.
00:26
So we see that this is going to be a shift of the graph of f down by two units.
00:44
So in particular, you know that a negative thing, or i guess just negative 2, to 1 on the x -axis, we're shifting it's down 2 units.
00:56
So this is going to move down 2 over here, and this will move down to 0.
01:05
So you have a line that's parallel, and then between 1 and 2, we're going to have the value of 0 for the function, and then between two and three, we're going to have another downward slope and line just parallel to the one above it.
01:32
And this is, we'll call this y1.
01:40
So this is function y1 in part b.
01:51
We're given that y is the function f of x minus 2.
01:58
You see that this is a shift of the graph of f by two units and because it's negative 2 to the right.
02:22
I'm going to draw this graph and you bring up.
02:26
I'm going to call it y2 and we'll see that going back to original graph f that means from x equals negative 2 to x equals 1 or to shift this line over to another parallel line.
03:02
So in particular, at x equal 1, we have then f of x is 2, so y2 of 3 is going to be 2, down one slope and line.
03:21
And the domain of this function is going to be starting at 0 instead of at negative 2.
03:30
That's part of it.
03:32
And then we see from x equals 3, we're going to have the same as the graph of f from one on.
03:50
So particular, we're going to move over one unit, and then we'll move over, down, one unit.
04:05
And this is going to be y2.
04:12
In part c, we're given that y is the function negative 2, f of a very 1 .5 .2.
04:21
This is a two different transformations.
04:27
So we have first stretch a graph vertically by a factor of two, then we reflect graph about the x -axis because of the negative signs.
05:09
And we'll call this one y3, draw it in blue.
05:13
So first things first, you want to take the function of that and stretch it.
05:26
So in particular, picking a few points here, we have at x equals negative 2, f of x is negative 1.
05:38
And so we're going to stretch that down to negative 2.
05:43
Then flip it across the axis so that the point becomes two, the value is two.
05:52
And because of reflection, we are drawing here the value that is two, so we double that, get four, and then flip it.
06:33
And we'll connect these two points by line.
06:38
And then from x equals 1, x equals 2, we have the f is 2.
06:45
So we double that, we get 4 again.
06:47
And then you flip it, we get another straight line there...