How to graph a quadratic function on a graphing calculator?

How does one determine if the graph of a trigonometric function is symmetrical about the y-axis, symmetric about the origin, or neither, without using a graphing calculator?

• Some examples: [math] f(x) = \frac{1}{1 - \sin (x)} [/math] [math] f(x) = \frac{1}{1 - \sin^2 (x)} [/math] [math] f(x) = \sin (x) + \cos (x) [/math]

• Answer:

  If a function is symmetric across the y-axis, then: [math]f(-x) = f(x)[/math] This is because if a function is symmetric across the y-axis, [math]f(x)[/math] and [math]f(-x)[/math] should give the same value. If a function is symmetric across the origin, then: [math]f(-x) = -f(x)[/math] This is because if a function is symmetric across the origin, then [math]f(-x)[/math] should give the opposite of [math]f(x)[/math] So plugging in [math]-x[/math] wherever an [math]x[/math] is present in the function and simplifying will help you determine. For the first function: Plug in [math]-x[/math] into [math]\sin x[/math]. This will give you [math]\sin (-x) = - \sin x[/math]. Plugging in [math]-\sin x[/math] as [math]\sin x[/math] will make the function equal to [math]f(x) = \dfrac{1}{1 + \sin x}[/math]. Since this new function is neither equal to nor the opposite of the original function (original function = function in description), the original function is neither symmetric across the y-axis nor symmetric across the origin. For the second function: We proceed similarly. Plugging in [math]-x[/math] into [math]\sin^2 x[/math] will give us [math](- \sin x)^2[/math] which is just [math]\sin^2 x[/math]. Thus, since this new function is equal to the original function, we have [math]f(x) = f(-x)[/math]. So the original function is symmetric across the y-axis. The third function is not symmetric to either the y-axis or origin, so I'll provide an example of a function symmetric to the origin. [math]f(x) = x^3[/math] is symmetric to the origin. Plugging in [math]-x[/math] into the function gives us [math]f(-x) = (-x)^3 = -x^3[/math]. Therefore, since [math]-x^3[/math] is the opposite of [math]x^3[/math], we have [math]f(-x) = -f(x)[/math], which means that this function is symmetric to the origin.