Transform y = 5cos3x to y = cos3x + 6

AI Thread Summary
Transforming the graph of y = 5cos(3x) to y = cos(3x + 6) involves a sequence of three transformations. First, the horizontal shift occurs by changing 3x to 3(x + 2), indicating a shift to the left by 2 units. Next, the vertical compression is applied by changing the amplitude from 5 to 1, which reduces the height of the graph. Finally, the function is adjusted to account for the new amplitude, resulting in y = 5cos(3x + 6). These transformations illustrate how changes in the function affect its graph's position and shape.
Calixto
Messages
16
Reaction score
0
Describe a sequence of transformations that would transform the graph of

y = 5cos3x into y = cos(3x + 6)
 
Physics news on Phys.org
The 5 in front of the cos(3x) (I assume it's cos(3x), be more clear next time) turned into a 1 in the second function. What does this mean?

The (3x) turned into a (3x + 6). What does this mean?

Recall the definitions of translations and expansions/compressions.
 
Remember that:

Let f(x), ang g(x), be two functions. Then if g(x)=f(x)+k, it means that g(x) is simply the function f(x) shifted up/down wards for k units.
g(x)=f(x+k), it means that g(x) is simply the function f(x) shifted horizontally for k units, either to the right or to the left, depending on the sign of the constant k.
g(x)=kf(x), it means that g is simply the function f, shrinked or extended(or how do you say it) vertically, depending whether |k|>1, or |k|<1.
g(x)=f(kx), is again the function f either extended, or shrinked horizontally, depending on the value of the constant k.

I hope this helps a lill bit.
 
Remember that any change before the given function is applied (in this case 5cos(3x)) is a change in x and any function after the function is a change in y.
changing 5cos(3x) to cos(3x+6) involves 3 changes:
1) Add 6 to 3x. That is, change 3x to 3x+ 6= 3(x+2) or x to (x+ 2). That's the first transformation.
2) change y= cos(3x+6) to y= 5cos(3x+6) or y to 5y. That's the second transformation.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top