Answer:
[tex]f(x)=(x+6)(x+6)[/tex]
Step-by-step explanation:
Let's find the roots of every function:
[tex]x(x-6)=0[/tex]
Expand the left side:
[tex]x^2-6x=0[/tex]
Using quadratic formula:
[tex]ax^2+bx+c=0\\x=\frac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex]
Therefore:
[tex]x=\frac{-(-6)\pm \sqrt{36-(4*1*0)} }{2*1} =\frac{6\pm \sqrt{36} }{2}=\frac{6\pm 6 }{2}[/tex]
Hence, the first function has two roots:
[tex]x=6\\or\\x=0[/tex]
Analizing the second function:
[tex](x-6)(x-6)=0[/tex]
It's easy to see that it has a repeated root at x=6
Now, the third function:
[tex](x+6)(x-6)=0[/tex]
Also, it's simple to conclude that it has two roots at x=6 and x=-6
Finally the fourth function:
[tex](x+6)(x+6)=0[/tex]
has a repeated root at x=-6, which satisfies the problem condition
Aditionally I attached the graph of every function, so you will be able to check the result easily.
