The first problem that was finished at 5:40 does have the mistake pointed out in the comments that the shift should be to the right because 2h(5 -x) = 2h[-(x-5)]. It is tricky when the input (x) is effected by shifts, stretches and reflections to know the correct order to make these transformations. I leave it to the reader to start with a well understood function like y = x^3 and discover the correct order. I comment only to say the original red graph of h(x) has all the info we need to check or find the correct graph of 2h(5-x). h has domain [-6,1] So for our new function 2h(5-x), we know the input 5-x must be between -6 and 1. So, -6 5 - x 1 which leads to 4 x 11 as the domain of 2h(5-x). Plugging in 4 we have 2h(5 - 4) = 2h(1) = 2(0) = 0 so the new function contains the point (4,0). Continue by plugging in 5,6,7,8,9,10, 11 to arrive at the new graph.
This video is actually wrong in its transformation of 2h(5-x). The horizontal shift should actually take place before the reflection. Inside he parentheses everything, including the order of transformations is reversed. Outside the parentheses things work normally and he order of transformations follows the order of operations. Also, if you are given h(x)=af(bx+h)+k then the point (x,y) on f(x) will transform to the point ((x-h)/b,ay+k) on the graph of h(x). This rule works for any a,b,h,k except b=0, in which case h(x) is a constant function that just makes a horizontal line.
knightofthebroho I agree, but an easier way to work with combinations is to start with your function in the form h(x)=a(b(x-c) +d. from here, just work your way left to right. If either a or b is negative, reflect prior to stretching or shrinking. So in the given example: 2h(-(x-5). Stretch the "y" by 2, reflect all points over the the y-axis then shift all points right 5 units.
The first problem at 1 minute is worked incorrectly. You need to factor the interior to correctly identify the reflection and horizontal transformation. So f(x)=2h(-1(x-5)) produces a vertical stretch by 2, a reflection over y-axis and shift right 5.
@@MySecretMathTutor you use an interesting approach and I like it. So I guess since the inside "does the opposite" that explains why you do the operations in opposite order?
@@MySecretMathTutor I've always taken a slightly different approach to transformations but similar. We'd do the multiplication operations first (in & out) then do the addition operations (in and out). So apply reflections and expansions before you apply shifts.
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THANK YOU SO MUCH. Finally there's a video that doesn't show just one transformation at a time.
BLESS YOUR SOUL SIR ! GREETINGS FROM CANADA.. you deserve an actual award sir.
I got a silver play button! :^D
You literally save my life, thanks a lot
The first problem that was finished at 5:40 does have the mistake pointed out in the comments that the shift should be to the right because 2h(5 -x) = 2h[-(x-5)]. It is tricky when the input (x) is effected by shifts, stretches and reflections to know the correct order to make these transformations. I leave it to the reader to start with a well understood function like y = x^3 and discover the correct order.
I comment only to say the original red graph of h(x) has all the info we need to check or find the correct graph of 2h(5-x). h has domain [-6,1] So for our new function 2h(5-x), we know the input 5-x must be between -6 and 1. So, -6 5 - x 1 which leads to 4 x 11 as the domain of 2h(5-x). Plugging in 4 we have 2h(5 - 4) = 2h(1) = 2(0) = 0 so the new function contains the point (4,0). Continue by plugging in 5,6,7,8,9,10, 11 to arrive at the new graph.
This video is actually wrong in its transformation of 2h(5-x). The horizontal shift should actually take place before the reflection. Inside he parentheses everything, including the order of transformations is reversed. Outside the parentheses things work normally and he order of transformations follows the order of operations.
Also, if you are given h(x)=af(bx+h)+k then the point (x,y) on f(x) will transform to the point ((x-h)/b,ay+k) on the graph of h(x). This rule works for any a,b,h,k except b=0, in which case h(x) is a constant function that just makes a horizontal line.
knightofthebroho I agree, but an easier way to work with combinations is to start with your function in the form h(x)=a(b(x-c) +d. from here, just work your way left to right. If either a or b is negative, reflect prior to stretching or shrinking. So in the given example: 2h(-(x-5). Stretch the "y" by 2, reflect all points over the the y-axis then shift all points right 5 units.
You betcha :-)
The first problem at 1 minute is worked incorrectly. You need to factor the interior to correctly identify the reflection and horizontal transformation. So f(x)=2h(-1(x-5)) produces a vertical stretch by 2, a reflection over y-axis and shift right 5.
@@MySecretMathTutor you use an interesting approach and I like it. So I guess since the inside "does the opposite" that explains why you do the operations in opposite order?
@@MySecretMathTutor I've always taken a slightly different approach to transformations but similar. We'd do the multiplication operations first (in & out) then do the addition operations (in and out). So apply reflections and expansions before you apply shifts.
That's how I've been able to keep it strait in my head, though I guess this video shows I still mix it up from time to time. :^D
okay but what do I do with the -1 on the inside
also what if the inside is more than 1/-1
Very good review, helped alot
perfectly explained. But i wish u could have squeezed a absolute value problem in there.
The first time I check out one of your videos, you have a major conceptual mistake on a key topic.
It does happen. Just let me know and I can mark it down in the errata, or if especially bad, I'll just re shoot the video. :^D
@@MySecretMathTutor Roger that. All the best to you. Keep helping the kids!