Need help explaining these transformations that take place in this function Y=3sinx to 3sin1/2(x-pi/2)
Is the new function \(y=3\sin(\frac{1}{2}(x-\pi/2))\)?
If it is, then you just have a shift right and horizontal stretch.
yep its a new functions im pretty sure and how muc of a shift to the right and horizontal shift?
Well lets think about it. If I have a function \(y=f(x)\), then what can I do to \(f(x)\) to make it such that all x's have to be, say, 3 larger than they actually are to have the same value (shift right by 3)?
yea
What do you mean 'yea'? I asked you a question.
oh and also the question ask for what the amp is and period and alsophase shift and what is the 1/4 wave?
and yes it has to be by 3
Calm down! we're not done with the first one. I AM ASKING YOU, if you have the function \(y=f(x)\), how do you shift right by 3?
f(-3)
no that gives the value of y AT -3. But you're close. Generalize for all x's.
x= -3
so all x's move over 3 units right
yes x-3 not = -3
so now we have f(x-3), which is a right shift by 3. What if I wanted to stretch all x values by a factor of 5? I need all values to be 5 times as big, just to equal the original f(x-3).
x^5 or x*5
Which one do you think. If I want 5x to be the same as x, then what do I multiply the 5 x by?
if you want the 5x to be the same as x? x1?
yes. you get it by multiplying by 1/5. if I transform f(x) with f(x 1/5) then I will have a horizontal stretch by a factor of 5, since x values need to be five times as large to equal what they were in f(x).
ok ic that
good. Then if I want a horiz shift of 3 and horiz stretch of 5 to f(x), what does it become?
3/5?
or is it 5/3?
NO! did i just get nothing done? f(x) - shift horizontally -> f(x-3) f(x-3) - stretch horizontally -> f((x-3) (1/5))
now, if I let f(x) = 3sin x, then I shifted horizontally by a and stretched horizontally by 1/b, then what does it look like now?
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