How do you find the important points to graph f(x)=x39x2+27x26?

1 Answer
May 12, 2017

See explanation...

Explanation:

Given:

f(x)=x39x2+27x26

I can see that this is quite similar to (x3)3. Let's expand that:

(x3)3=x3+3(x2)(3)+3(x)(3)2+(3)3

(x3)3=x39x2+27x27

So we find:

f(x)=(x3)3+1

Now the sum of cubes identity can be written:

a3+b3=(a+b)(a2ab+b2)

So we find:

f(x)=(x3)3+1

f(x)=(x3)3+13

f(x)=((x3)+1)((x3)2(x3)+1)

f(x)=(x2)(x26x+9x+3+1)

f(x)=(x2)(x27x+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=x3 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]}