Question

Which function has only one x-intercept at (−6, 0)?

f(x) = x(x − 6)
f(x) = (x − 6)(x − 6)
f(x) = (x + 6)(x − 6)
f(x) = (x + 6)(x + 6)

Answer 1

`f(x)=(x+6)(x+6)`

Explanation:

Let's find the roots of every function:

`x(x-6)=0`

Expand the left side:

`x^2-6x=0`

Using quadratic formula:

`ax^2+bx+c=0\\x=(-b\pm √(b^2-4ac) )/(2a)`

Therefore:

`x=(-(-6)\pm √(36-(4*1*0)) )/(2*1) =(6\pm √(36) )/(2)=(6\pm 6 )/(2)`

Hence, the first function has two roots:

`x=6\\or\\x=0`

Analizing the second function:

`(x-6)(x-6)=0`

It's easy to see that it has a repeated root at x=6

Now, the third function:

`(x+6)(x-6)=0`

Also, it's simple to conclude that it has two roots at x=6 and x=-6

Finally the fourth function:

`(x+6)(x+6)=0`

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.

Answer 2

The function f(x) = (x + 6)(x + 6) has only one x-intercept at (-6, 0).