Dada la función f(x)=1+6Sen(2x+π/3) . Halle: Período, amplitud y desfase (1.5 puntos)
Dominio y rango de la función (1.5 puntos)
Grafique la función trigonométrica (2 puntos)

\displaystyle f(x)=\frac{10}{169}\left(x-8\right)^2-1f(x)=16910(x−8)2−1

Step-by-step explanation:

We can use the vertex form of a quadratic, which is given by:

\displaystyle f(x)=a(x-h)^2+kf(x)=a(x−h)2+k

Where (h, k) is the vertex.

Since we are given that the vertex is (8, -1), h = 8 and k = -1. Substitute:

f(x)=a(x-8)^2-1f(x)=a(x−8)2−1

Next, we are given that the parabola passes through the point (-5, 9). So, when x = -5, y = 9:

9=a((-5)-8)^2-19=a((−5)−8)2−1

Solve for a:

9=a(-13)^2-19=a(−13)2−1

So:

\displaystyle a=\frac{10}{169}a=16910

So, the equation of our parabola is:

\displaystyle f(x)=\frac{10}{169}\left(x-8\right)^2-1f(x)=16910(x−8)2−1