Question

Help me plss, i need this done by today

Answer 1

`~~~~~~~~~~~~\textit{function transformations} \\\\\\ f(x)=Asin(Bx+C)+D \qquad \qquad f(x)=Acos(Bx+C)+D \\\\ f(x)=Atan(Bx+C)+D \qquad \qquad f(x)=Asec(Bx+C)+D \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks}\\ ~~~~~~\textit{horizontally by amplitude } A\cdot B\\\\ \bullet \textit{ flips it upside-down if }A\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if }B\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis}`

`\bullet \textit{ horizontal shift by }(C)/(B)\\ ~~~~~~if\ (C)/(B)\textit{ is negative, to the right}\\\\ ~~~~~~if\ (C)/(B)\textit{ is positive, to the left}\\\\ \bullet \textit{vertical shift by }D\\ ~~~~~~if\ D\textit{ is negative, downwards}\\\\ ~~~~~~if\ D\textit{ is positive, upwards}\\\\ \bullet \textit{function period or frequency}\\ ~~~~~~(2\pi )/(B)\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\ ~~~~~~(\pi )/(B)\ for\ tan(\theta),\ cot(\theta)`

with the template above in mind, let's check the graph of that function, it has a midline at y = 0, namely the x-axis, so it has the same midline as sine, we'll use sine, because sine graph starts at 0 and then it moves up and then down, this one starts at 0, reason why, I guess we could also use a -cosine, which has the same effect, anyhow, let's use sine.

from the midlne up, we have a peak at 1.5 or 3/2 which our Amplitude.

we can also see that from 0 to 8, is the graph, after that it simply repeats, so we can say that the period of that one is 8, thus
`\stackrel{period}{8}=\cfrac{2\pi }{B}\implies B=\cfrac{2\pi }{8}\implies B=\cfrac{\pi }{4}`

and since the midline is same as sin(θ) that means it doesn't have a vertical shift, so our D = 0, and since it's beginning at x = 0, it doesn't have any horizontal shift from the sin(θ) either, so C = 0, thus we can write it as

`y = \stackrel{A}{\cfrac{3}{2}}sin\left( \stackrel{B}{(\pi )/(4)}x-\stackrel{C}{0} \right)+\stackrel{D}{0}\implies y = \cfrac{3}{2}sin\left( (\pi )/(4)x \right)`

Check the picture below.