Hello thisisamazing Originally Posted by
thisisamazing ok for example:
Sin(x+pi/2)
or
-1/2cos(4x+pi)
First, try thinking about the value of $\displaystyle x$ that makes the expression in the brackets zero. This will tell you how far the 'basic' graph has been moved. So for example, in$\displaystyle y = \sin(x+\pi/2)$
the expression $\displaystyle x +\pi/2$ is zero when $\displaystyle x = -\pi/2$. So, instead of 'starting' at $\displaystyle (0,0)$ the sine graph has been moved to 'start' at $\displaystyle (-\pi/2,0)$; i.e. it has been moved a distance of $\displaystyle \pi/2$ to the left.
(You'll see that I've put the word 'start' in quotes. That's because, of course, the graph doesn't actually start there; it extends infinitely far in each direction. But the cycle of the graph that we usually look at will start there.)
In the second example you give, $\displaystyle y = -\tfrac12\cos(4x+\pi)$: $\displaystyle 4x + \pi =0$
when$\displaystyle x=-\pi/4$
So the 'basic' cosine graph has been moved $\displaystyle \pi/4$ to the left to 'start' at $\displaystyle x=-\pi/4$.
However, a couple of other things have happened here as well:
- $\displaystyle x$ is multiplied by $\displaystyle 4$, which means that things happen $\displaystyle 4$ times as fast along the $\displaystyle x$-axis as they will on the 'basic' graph. So instead of requiring values of $\displaystyle x$ from $\displaystyle 0$ to $\displaystyle 2\pi$ to make a complete cycle, you'll only need values from $\displaystyle 0$ to $\displaystyle \pi/2$.
- The cosine expression has then been multiplied by $\displaystyle -\tfrac12$. This, in turn, does two things:
- The minus sign flips the graph over, reflecting it in the $\displaystyle x$-axis.
- The factor of $\displaystyle \tfrac12$ reduces the $\displaystyle y$-values to one-half of their original values, 'squashing' the graph down, so that, instead of varying between $\displaystyle -1$ and $\displaystyle +1$, it will now vary between -0.5 and $\displaystyle +0.5$.
Does that help to make it clear?
Grandad