# How do you graph #f(x)=x^4-4x^3+x^2+6x#?

##### 1 Answer

I would recommend starting by factoring. We can instantly factor out an

#f(x) = x(x^3 - 4x^2 + x + 6)# .

Now, use the rational root theorem to determine the factors. For a polynomial

#"possible factors" = "(+-1, +-2, +-3, +-6)/(+- 1)#

#"possible factors" = +-1, +-2, +-3, +-6#

I usually start at

The remainder theorem will confirm which of those numbers are factors. If

#f(1) = 1(1^3 - 4(1)^2 + 1 + 6) = 4" "color(red)(xx)#

#f(-1) = 1((-1)^3 - 4(-1)^2 - 1 + 6) =0" "color(green)(√)#

We have a factor. Now use either synthetic or long division to divide

So, we can now write

Hence,

There will be four x-intercepts, at

They all have multiplicity

We will have to repeat the above process to find the zeroes of the derivative, which will be the critical points.

Once you have done this, you should obtain

There will be critical points at

#0 = 2x^2 - 4x - 3#

#x = (-(-4) +- sqrt((-4)^2 - (4 * 2 * -3)))/(2 * 2)#

#x= (4 +- sqrt(40))/4#

#x = (4 +- 2sqrt(10))/4#

#x = 1 +- 1/2sqrt(10)#

Test to the left and right of all of these points, by taking test points and plugging them into the derivative. For example, If they are negative, then the function is decreasing in that interval. If the function is decreasing *to the left* of the critical point, and increasing to the right, you most certainly have a local minimum, for example.

Doing this, you will find that there are local minimums at

The final step is checking end behaviour. This will be determined by the leading term in the initial polynomial, or

This proves that *absolute minimums*, that is the smallest y-value on the function. The function never has an absolute maximum. Here is the graph of the function:

Hopefully this helps!