Question

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

Answer 1

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

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

Now, use the rational root theorem to determine the factors. For a polynomial `f(x)= ax^n + bx^m + ... + p`, the possible factors are given by `"factors of p"/"factors of q"`.

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

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

I usually start at `1` and work my way through the possibilities.

The remainder theorem will confirm which of those numbers are factors. If `x =a` is a factor of `f(x)`, then `f(a) = 0`.

`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 `x^3 - 4x^2 + x + 6` by `x + 1`.

So, we can now write `f(x) = x^4 - 4x^3 + x^2 + 6x` as `f(x) = x(x + 1)(x^2 - 5x + 6)`. Now, we can factor `x^2 - 5x + 6` as `(x - 3)(x - 2)`.

Hence, `f(x) = x(x + 1)(x - 2)(x - 3)`

There will be four x-intercepts, at `x = 0`, `x = -1`, `x = 2` and `x = 3`.

They all have multiplicity `1`, or they are of degree `1`, so it will be a smooth cut over the x-axis. Next I would use a little bit of calculus to test for critical values + local/absolute max/mins.

`f'(x) = 4x^3 - 12x^2 + 2x + 6`

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 `2(x- 1)(2x^2 - 4x - 3)`.

There will be critical points at `x = 1` and the solutions of `2x^2 - 4x - 3`. Solving using the quadratic formula, we have:

`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 `x =1 +- 1/2sqrt(10)` and a local maximum at `x = 1`.

The final step is checking end behaviour. This will be determined by the leading term in the initial polynomial, or `x^4`. The following table explains best.

`x^4` is of even degree and is positive.

This proves that `x = 1 +- 1/2sqrt(10)` aren't local minimums but instead 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!