How do you graph #f(x)=2-x^3# using zeros and end behavior?

1 Answer
Oct 2, 2017

Left end is upwards, right end is downwards, and the graph crosses the x-axis at #(root(3)(2),0)#.

Explanation:

This function #f(x)# is a polynomial of degree 3. Begin by rearranging the function in decreasing order of exponents:

#f(x) = -x^3+2#

Since this is a degree 3 polynomial, the odd degree (3) tells us that the left and right ends of the graph will point in opposite directions.

The leading coefficient is -1, which indicates that the right end of the graph will point downwards. (Hint: the negative leading coefficient means the right tail points towards the negative #y# values, or down.) Given we know the left tail is opposite to the right, we now know the left tail will rise/point upwards.

Next we should check for both #x#- and #y#-intercepts. #x#-intercepts happen when #y = 0#, or:

#-x^3+2 = 0#
#x^3 = 2#
#x = root(3)(2) ~~1.26#

#y#-intercepts happen when #x = 0#, or:

#y = -(0^3)+2#
#y = 2#

The rough sketch using just this information should behave following these patterns:

  • The graph comes down from the top of the graph paper on the left side of the y-axis.
  • The graph crosses over the y-axis at (0,2), and heads towards the x-axis.
  • The graph crosses over the x-axis at #(root(3)(2),0)# heading downwards.
  • The graph exits off the bottom of the graph paper to the right of #x=root(3)(2)#

Here's how it looks:

graph{2-x^3 [-4.656, 5.21, -0.705, 4.228]}