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

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Answer 1 · 2016-11-01T12:00:47

See Explanation

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

$color(brown)("Consider the case "x<0)$

$x^2>0 " so "-x^2<0$

In this context $x<0" so "x^3<0$

But in this case

$|x^3|>|x^2|" so "x^3" has the greater influence "->x^3-x^2<0$

$color(blue)(lim_(x->-oo) x^3-x^2->-oo )$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(brown)("Consider the case "x>0)$

$x^2>0 " so "-x^2<0$

In this context $x>0" so "x^3>0$

$|x^3|>|x^2|" so "x^3" has the greater influence "->x^3-x^2>0$

$color(blue)(lim_(x->+oo) x^3-x^2->+oo)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(brown)("General shape y-intercept point")$

y-intercept at $x=0 -> y=(0)^3-(0)^2+4=4$

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.

Tony B

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