![]() ![]() When x is equal to four, four minus three is one,Ībsolute value of one is indeed, is indeed one. So when x is equal to three, three minus three is zero, absolute value of that is zero. If I say y is equal to theĪbsolute value of x minus three, well, let's try it out. Value of something and so you say, okay, if x is three, how do I make that equal to zero? Well, I can subtract three from it. Looks like an absolute value so it's going to have the form, y is equal to the absolute Going to switch signs and so we say, okay, well, this Needs to evaluate out to zero and this is where it's So at this point right over here, we know that our function, we know that our equation Switch in direction here of this line and so you see the same thing Instead of taking anĪbsolute value of a negative, you're now taking the absolute value as you cross this point of a positive and that's why we see a Thing that happens here is that you switch signs So that's what we first wannaįigure out the equation for and so how would we think about it? Well, one way to think about it is, well, something interesting is happening right over here at x equals three. Three to the right, it would look like, it would look like this. Shift three to the right and think about how that The right and four up? Alright, now let's do this together. Video like we always say and figure out what would the equation be if you shift three to Of this graph if we shift, if we shift three to the right and then think about how that will change if not only do we shift three to the right but we also shift four up, shift four up, and so once again pause this ![]() Gonna first think about what would be the equation What I wanna do in this video is think about how theĮquation will change if we were to shift this graph. If you take x is equal to negative two, the absolute value of To absolute value of x which you might be familiar with. This right over here is the graph of y is equal |5+a|=0 So what does a have to be here? obviously -5 so that tells us |x-5| has the 0 point now at x=5, or in other words the graph was moved 5 places to the right. We know we have to add or subtract something inside to make it happen. ![]() My suggestion is to think backwards with an answer and what youw ould need to change. If you are asking why it moves like that when you add or subtract then that is a little more tricky to answer. if you add you go left, so |x+3| goes to the left 3. Once you find the inside of the function you just need to subtract a number from the variable to move right. inside the function is inside the absolute value bars. Specifically to move a graph to the right you need to determine the inside of the function. You can move it up, down, left, and right. Why it moves 3 to the right is because you can move graphs around. When you graph the absolute value function it makes a sudden sharp turn when you get to 0, which in other words is saying when you SWITCH SIGNS fro the negative numbers to non negative, the graph turns. |-1| = 1 and all other negative numbers are turned to positive. See Problem 1c) below.When he mentions switching signs he means what is inside of the abslute value signs. The argument x of f( x) is replaced by − x. And every point that was on the left gets reflected to the right. Every point that was to the right of the origin gets reflected to the left. Every y-value is the negative of the original f( x).įig. Its reflection about the x-axis is y = − f( x). Only the roots, −1 and 3, are invariant.Īgain, Fig. And every point below the x-axis gets reflected above the x-axis. Every point that was above the x-axis gets reflected to below the x-axis. The distance from the origin to ( a, b) is equal to the distance from the origin to (− a, − b).į( x) = x 2 − 2 x − 3 = ( x + 1)( x − 3).įig. If we reflect ( a, b) about the x-axis, then it is reflected to the fourth quadrant point ( a, − b).įinally, if we reflect ( a, b) through the origin, then it is reflected to the third quadrant point (− a, − b). It is reflected to the second quadrant point (− a, b). C ONSIDER THE FIRST QUADRANT point ( a, b), and let us reflect it about the y-axis. ![]()
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