Here's the question you clicked on:
sabika13
Write an equation this is equivalent to f(x)=-5sin(x-pi/2) - 8 using trig.function of x
ok so you have \[f(x)=-5\sin\left(x-\frac{\pi}{2}\right)-8\] For now just focus on \[g(x)=\sin\left(x-\frac{\pi}{2}\right)\] what does this do to the graph?
where \(x=\pi/2\), what is \[\sin\left(x-\frac{\pi}{2}\right)\]also brb. just gotta hand in something quick...
okay.. will it be sin(x-pi/2)=0??
i dont really understand the question itself
sorry open study died on me.
You are correct that \(g(\pi/2)=0\) You can repeat this with some other key x-values \(x=\pi/2,\quad x=\pi ,\quad x=0 \) and find that what you get is \[\sin\left(x-\frac{\pi}{2}\right)=-\cos\pi\]
you will eventually just memorize these phase shifts, but if you ever forget on a test, you can always make a quick sketch using the method I explained. So now remember that we called \[g(x)=\sin\left(x-\frac{\pi}{2}\right)=-\cos x\] Going back to the original function you have \[f(x)=-5\sin\left(x-\frac{\pi}{2}\right)-8\]\[f(x)=-5g(x)-8\]So the hard part is already done. The 8 is all by itself so it just shifts the up and down, and the -5 just makes the amplitude of the wave greater. All you have to do now is plug in the expression you worked out for g(x). \[f(x)=-5(-\cos x)-8\]\[f(x)=5\cos x -8\]
Oops made a mistake, in the second last post. It should say : You are correct that \(g(π/2)=0\) You can repeat this with some other key x-values \(x=π/2,\quad x=π,\quad x=0\) and find that what you get is \[\sin\left( x−\frac{\pi}{2}\right)=−\cos x\]
sorry i fell alseep -_- .. accoriding the the cofunction identities isnt it : cosx=sin(pi/2-x)? so using that you can get -cosx=sin(x-pi/2) (sorrydont understand)