Question

How do you find the important points to graph `f(x) = x^3 - 9x^2 + 27x - 26`?

Answer 1

See explanation...

Explanation:

Given:

`f(x) = x^3-9x^2+27x-26`

I can see that this is quite similar to `(x-3)^3`. Let's expand that:

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

`color(white)((x-3)^3) = x^3-9x^2+27x-27`

So we find:

`f(x) = (x-3)^3+1`

Now the sum of cubes identity can be written:

`a^3+b^3 = (a+b)(a^2-ab+b^2)`

So we find:

`f(x) = (x-3)^3+1`

`color(white)(f(x)) = (x-3)^3+1^3`

`color(white)(f(x)) = ((x-3)+1)((x-3)^2-(x-3)+1)`

`color(white)(f(x)) = (x-2)(x^2-6x+9-x+3+1)`

`color(white)(f(x)) = (x-2)(x^2-7x+13)`

Note that `f(0) = -26`

So we have:

• `f(x)` has `x` intercept `(2, 0)`

• `f(x)` has `y` intercept `(0, -26)`

• `f(x)` is like `y = x^3` but shifted right by `3` units and up by `1` unit.

Putting this all together we find `f(x)` looks like this:

graph{(y-(x^3-9x^2+27x-26))(6x^2+(y+26)^2-0.06)(6(x-2)^2+y^2-0.06)(6(x-3)^2+(y-1)^2-0.06) = 0 [-7.375, 12.625, -34, 18]}