Question

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

Answer 1

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]}