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

Oct 2, 2017

Left end is upwards, right end is downwards, and the graph crosses the x-axis at $\left(\sqrt[3]{2} , 0\right)$.

#### Explanation:

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

$f \left(x\right) = - {x}^{3} + 2$

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:

$- {x}^{3} + 2 = 0$
${x}^{3} = 2$
$x = \sqrt[3]{2} \approx 1.26$

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

$y = - \left({0}^{3}\right) + 2$
$y = 2$

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 $\left(\sqrt[3]{2} , 0\right)$ heading downwards.
• The graph exits off the bottom of the graph paper to the right of $x = \sqrt[3]{2}$

Here's how it looks:

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