How do you graph $y = 1 + 2 sec x$ $y=1+2secx$ ?

Question

3552 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-29T02:23:01

As below.

Standard form is $y = a sec (B x - C) + D$ $y = a sec (Bx -C) + D$

Given equation is $y = 2 sec (x) + 1$ $y = 2 sec (x) + 1$

$A m p l i t u d e = None, for secant function$ $Amplitude = " None, for secant function"$

$Period = \frac{2 π}{| B |} = 2 \frac{π}{1} = 2 π$ $"Period " = (2pi) / |B| = 2pi / 1 = 2pi$

$" Phase shift = (-C) / B = 0$

$"Vertical Shift " = D = 1$

graph{2 sec(x) + 1 [-10, 10, -5, 5]}

How do you graph y=1+2secxy=1+2secxy=1+2secx?