Answer:
The way you would sketch the graph is slowly increasing, staying about the same, and then go up a little bit.
Good luck!
The domain of f(x) = 1 + tan^2 x is all real numbers.
The range of f(x) = 1 + tan^2 x is 1 ≤ f(x) ≤ 2
Let the shorter leg = a.
The longer leg, b = a + 4.
The hypotenuse, c = a + 8
a^2 + b^2 = c^2
a^2 + (a + 4)^2 = (a + 8)^2
a^2 + a^2 + 8a + 16 = a^2 + 16a + 64
a^2 - 8a - 48 = 0
(a - 12)(a + 4) = 0
a - 12 = 0 or a + 4 = 0
a = 12 or a = -4
a = -4 is discarded because it is negative.
a = 12
b = 16
c = 20
The lengths are 12 m, 16 m, and 20 m.
<u> Solution-</u>
The given function is,






Therefore, at x=0, -1, 1 , f(x) will be 0 . Hence, 0, -1 ,1 are the x-intercepts.
Plotting the graph on desmos, the graph will be as in the attachment.