# Verify if the function is odd or even? y=17x^3-12x^2

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### 2 Answers

A function is odd if f(-x) = -f(x) and even if f(-x) = f(x)

Here we have f(x) = y= 17x^3 - 12x^2

f(-x) = -17x^3 - 12x^2

f(x) = 17x^3 - 12x^2

-f(x) = -17x^3 + 12x^2

So we see that f(-x) = -17x^3 - 12x^2 is neither equal to f(x) nor equal to -f(x).

**Therefore the function is neither even nor odd**.

A function is even if f(-x) = f(x).

In other words, plugging in a number will be the same as plugging in the negative value of the same number. The function is not changing.

We'll analyze the given function, replacing each x by -x.

f(-x) = 17(-x)^3 - 12(-x)^2

We'll compute raising -x to the 3rd and 2nd powers and we'll get:

(-x)^3 = (-x)(-x)(-x) = x^2*(-x) = -x^3

f(-x) = -17x^3 - 12x^2

So we can see that:

f(-x) is not equal to f(x) which means that the function f(x) is not an even function.

We'll check if the function is odd;

f(-x) = -f(x)

f(-x) = -17x^3 - 12x^2

If we'll factorize by -1 we'll get:

f(-x) = -(17x^3 + 12x^2)

The expression inside brackets is not the function f(x).

**The given function is nor odd neither even function.**