# Find the horizontal and vertical asymptotes of the graph of the function f(x) = x^3-6x^2+3x+10, if any.Find also the extreme points for and points of inflection of the function, if any. Make a...

Find the horizontal and vertical asymptotes of the graph of the function f(x) = x^3-6x^2+3x+10, if any.

Find also the extreme points for and points of inflection of the function, if any. Make a study of the concavity and growth of all points of function and then the sketch of the graph.

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Since the function is a polynomial, there are no asymptotes.  However, we expect that there are two extrema (since the polynomial is degree 3) and that there is one inflection point.  To find those points, we need to take derivatives.

`f'(x)=3x^2-12x+3`   now factor

`=3(x^2-4x+1)`

`f''(x)=6x-12`   again need to factor

`=6(x-2)`

The inflection point is at `f''(x)=0` which is when `x=2` . The y-value is `f(2)=3(2^2)-12(2)+3=-9`  so the inflection point is (2,-9).

The extrema are found at `f'(x)=0` , which can be found using the quardatic formula.  Then

`x={4+-sqrt{16-4(1)}}/2=1/2(4+-2sqrt3}=2+-sqrt3`

The two extreme points are `x approx 4.7` and `x approx 0.3` .

By the second derivative test, when x=0.3, `f''<0` , so the point is a local maximum.  Similarly, when x=4.7, `f''>0` , so the point is a local minimum.

The function is concave up when `f''>0` which is `x>2` .  The function is concave down when `f''<0` which is `x<2` .

The function is increasing when `f'>0` , which is on the intervals `(-infty,2-sqrt3)` and `(2+sqrt3,infty)` .  The function is decreasing when `f'<0` , which is on the interval `(2-sqrt3,2+sqrt3)` .

Combining this information gives the graph: