Given the equation:
![y=2\cos (4x)+1](https://img.qammunity.org/2023/formulas/mathematics/college/mzmpcag4wkb3ndi7w0zm8t9hqzwlq84c6l.png)
Let's graph the equation.
Apply the form:
![y=a\cos (bx-c)+d](https://img.qammunity.org/2023/formulas/mathematics/college/y0nx2dl226x4tosw3hfl9r7i6gfx7qfamo.png)
Where:
a = 2
b = 4
c = 0
d = 1
Let's determine the amplitude, period, phase shift and vertical shif.
Amplitutde = a = 2
To find the period, we have:
![(2\pi)/(b)=(2\pi)/(4)=(\pi)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/bcidhpm3zkkcpq6kh6wns9sus8qxuxkig1.png)
Phase shift:
![(c)/(b)=(0)/(4)=0](https://img.qammunity.org/2023/formulas/mathematics/college/pqnfsc350x2v6nenczzkogottrdr67iwuy.png)
Vertical shift, d = 1
Find few points of the equation on a graph.
We have:
When x = 0
y = 2 cos (4(0))+1
y = 2 cos (0)+1
y = 2 (1) + 1
y = 2 + 1
y = 3
When x = π/8
y = 2 cos (4(π/8)) + 1
y = 2 cos(π/2) + 1
y = 2 (0) + 1
y = 0 + 1
y = 1
When x = π/4
y = 2 cos(4(π/4))+1
y = 2 cos(π) + 1
y = 2 (-cos(0)) + 1
y = 2(-1) + 1
y = -2 + 1
y = -1
WHen x = π/2
y = 2 cos(4(π/2)) + 1
y = 2 cos(2π) + 1
y = 2 cos(0) + 1
y = 2(1) + 1
y = 2 + 1
y = 3 =
Thus, we have the points:
(0, 3), (π/8, 1), (π/4, -1), (π/2, 3)
The graph of the equation is attached below: