Question

Sketch two periods of the graph of the function h(x)=3sec(π4(x+3))

. Identify the stretching factor, period, and asymptotes. Enter the exact answers. Stretching factor =3 Period: P=8.Enter the asymptotes of the function on the domain [−P,P] .To enter π
, type Pi.
The field below accepts a list of numbers or formulas separated by semicolons (e.g. 2;4;6 or x+1;x−1 ). The order of the list does not matter.
Asymptotes: x=

Answer 1

To sketch the graph of h(x) = 3sec(π/4(x+3)), identify the stretching factor, period, and asymptotes.

Step-by-step explanation:

To sketch the graph of the function h(x) = 3sec(π/4(x+3)), we need to identify the stretching factor, period, and asymptotes.

The stretching factor is 3, which means the graph will be vertically stretched by a factor of 3.

The period can be found by calculating the distance between consecutive asymptotes. In this case, the period is 8 units.

The graph will have vertical asymptotes at values of x that make the secant function undefined. To find these values, we can set the expression inside the secant function equal to the asymptote value and solve for x:

π/4(x+3) = (2n + 1)π/2, where n is an integer

Simplifying the equation, we get:

x+3 = (4n + 1)/2, where n is an integer

Solving for x, we get:

x = (4n - 1)/2 - 3, where n is an integer

So, the asymptotes are x = -2.5, x = -6.5, x = -10.5, ...