find the equation of a sine wave in the form y = a sin(bx + c) + d that is obtained by vertically stretching…
find the equation of a sine wave in the form y = a sin(bx + c) + d that is obtained by vertically stretching it by a factor of 10, shifting it downward by 8 units, having a period of 9 units, and shifting it to the right by 6 units. a = b = c = d = here is a desmos graph that may help. question help: written example post to forum
Answer
Explanation:
Step1: Determine the value of a
Vertical - stretch factor gives a. Since it is vertically stretched by a factor of 10, $a = 10$.
Step2: Determine the value of b
The period formula for $y=a\sin(bx + c)+d$ is $T=\frac{2\pi}{|b|}$. Given $T = 9$, then $9=\frac{2\pi}{|b|}$, so $b=\frac{2\pi}{9}$.
Step3: Determine the value of c
For a right - shift of $h$ units in $y = \sin(x)$, the transformation is $y=\sin(x - h)$. Here, $h = 6$, and for $y=a\sin(bx + c)+d$, we have $bx + c=b(x - 6)$. Expanding gives $bx + c=bx-6b$. Substituting $b=\frac{2\pi}{9}$, we get $c=- \frac{4\pi}{3}$.
Step4: Determine the value of d
A downward shift of $k$ units gives $d=-k$. Since it is shifted downward by 8 units, $d=-8$.
Answer:
$a = 10$ $b=\frac{2\pi}{9}$ $c=-\frac{4\pi}{3}$ $d=-8$