How do you determine whether #f(t)= t^2 + 2t + -3# is an odd or even function?

Question

Sadie Alexander · Accepted Answer

This function is neither odd or even.

To test whether a function is odd or even, substitute #-t# for #t#. #f(t) = (-t)^2 + 2(-t) + -3# #f(t) = t^2 - 2t - 3# If it's even, you would get the same equation as you started with. If it's odd, you would get the opposite signs in the equation that you started with. This function would be odd if substituting #-t# gave you #-t^2 - 2t + (-3)# You could also graph the function to check. It is even if the graph is symmetric with respect to the y-axis, and odd if the graph is symmetric with respect to the origin.

Sadie Alexander · Answer

undefined To determine whether $ f(t) = t^2 + 2t - 3 $ is an odd or even function, we examine its symmetry properties:

1. Odd Function: $ f(t) $ is odd if $ f(-t) = -f(t) $ for all $ t $ in the domain.
2. Even Function: $ f(t) $ is even if $ f(-t) = f(t) $ for all $ t $ in the domain.

We'll evaluate $ f(-t) $ and $ f(t) $ to see if they are equal or opposite to each other:

$$ f(-t) = (-t)^2 + 2(-t) - 3 = t^2 - 2t - 3 $$
$$ f(t) = t^2 + 2t - 3 $$

Since $ f(-t) $ is not equal to $ f(t) $, and they are not negatives of each other, the function $ f(t) = t^2 + 2t - 3 $ is neither odd nor even.