What is the zeros, degree and end behavior of #y=-2x(x-1)(x+5)#?

1 Answer
May 30, 2018

Answer:

Zeroes

#x \in \{-5, 0, 1\}#

Degree

Polynomial of third degree

End behaviour

#lim_{x\to+\infty}-2x(x-1)(x+5) = -\infty#

#lim_{x\to-\infty}-2x(x-1)(x+5) = +\infty#

Explanation:

Zeroes

This is very easy: the function is already written in its factorized form. So, if you want to solve

#-2x(x-1)(x+5)=0#

you are asking for a multiplication to be zero. A multiplication is zero if and only if at least one of its factors is zero, so the alternatives are

  • #-2x = 0 \iff x = 0#
  • #x-1 = 0 \iff x = 1#
  • #x+5 = 0 \iff x = -5#

Degree

Just by eyeballing the equation, you can tell this is a polynomial of degree three, since it's the multiplication of three degrees of degree one.

But just to be sure, let's do the actual multiplications:

#\color(red)(-2x(x-1))(x+5) = \color(red)((-2x^2+2x))(x+5) = -2x^3-10x^2+2x^2+10x = -2x^3-8x^2+10x#

End Behaviour

The end behaviour is a direct consequence of the degree. If we call any polynomial of even degree #f_{even}(x)# and any polynomial of odd degree #f_{odd}(x)#, the end behaviours will be (assuming the leading term is positive):

#lim_{x\to\pm\infty}f_{even}(x) = \infty#
#lim_{x\to\pm\infty}f_{odd}(x) = \pm\infty#

Since you have a minus sign in front of the polynomial, the limits will be inverted.