How do you find the important points to graph f(x)=x3−9x2+27x−26?
1 Answer
May 12, 2017
See explanation...
Explanation:
Given:
f(x)=x3−9x2+27x−26
I can see that this is quite similar to
(x−3)3=x3+3(x2)(−3)+3(x)(−3)2+(−3)3
(x−3)3=x3−9x2+27x−27
So we find:
f(x)=(x−3)3+1
Now the sum of cubes identity can be written:
a3+b3=(a+b)(a2−ab+b2)
So we find:
f(x)=(x−3)3+1
f(x)=(x−3)3+13
f(x)=((x−3)+1)((x−3)2−(x−3)+1)
f(x)=(x−2)(x2−6x+9−x+3+1)
f(x)=(x−2)(x2−7x+13)
Note that
So we have:
-
f(x) hasx intercept(2,0) -
f(x) hasy intercept(0,−26) -
f(x) is likey=x3 but shifted right by3 units and up by1 unit.
Putting this all together we find
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]}