Write a sine function that has an amplitude of 5, a midline of 3 and a period of 1/8.

Mathematics

1 answer:

vodomira [7]2 years ago

6 0

Answer:

y=5sin(16pix)+3

Step-by-step explanation:

Amp=5 means our curve is either y=5sin(bx+c)+d or y=-5sin(bx+c)+d.

y=sin(x) has period 2pi.

So y=sin(bx) has period 2pi/b.

We want 2pi/b=1/8.

Cross multiplying gives: 16pi=b

y=5sin(16pix+c)+d

d=3 since we want midline y=3.

y=5sin(16pix+c)+3

We can choose c=0 since we aren't required to have a certain phase shift.

y=5sin(16pix)+3

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Refer to the diagram below.

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$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-114)\pm\sqrt{(-114)^2-4(4)(392)}}{2(4)}\\\\x = \frac{114\pm\sqrt{6724}}{8}\\\\x = \frac{114\pm82}{8}\\\\x = \frac{114+82}{8} \ \text{ or } \ x = \frac{114-82}{8}\\\\x = \frac{196}{8} \ \text{ or } \ x = \frac{32}{8}\\\\x = 24.5 \ \text{ or } \ x = 4\\\\$