# How do you sketch the general shape of f(x)=x^3-x^2+4 using end behavior?

Nov 1, 2016

See Explanation

#### Explanation:

Consider end behaviour. That is when $x$ tends to extreme negative and extreme positive for this function.

$\textcolor{b r o w n}{\text{Consider the case } x < 0}$

${x}^{2} > 0 \text{ so } - {x}^{2} < 0$

In this context $x < 0 \text{ so } {x}^{3} < 0$

But in this case

$| {x}^{3} | > | {x}^{2} | \text{ so "x^3" has the greater influence } \to {x}^{3} - {x}^{2} < 0$

$\textcolor{b l u e}{{\lim}_{x \to - \infty} {x}^{3} - {x}^{2} \to - \infty}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Consider the case } x > 0}$

${x}^{2} > 0 \text{ so } - {x}^{2} < 0$

In this context $x > 0 \text{ so } {x}^{3} > 0$

$| {x}^{3} | > | {x}^{2} | \text{ so "x^3" has the greater influence } \to {x}^{3} - {x}^{2} > 0$

$\textcolor{b l u e}{{\lim}_{x \to + \infty} {x}^{3} - {x}^{2} \to + \infty}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{General shape y-intercept point}}$

y-intercept at $x = 0 \to y = {\left(0\right)}^{3} - {\left(0\right)}^{2} + 4 = 4$

The x-intercept is harder to determine but as you are only asked to sketch the 'general case' it is not important that you find it.