Question

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

Answer 1

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

Answer 2

Answer:

$f(x)=(x+6)(x+6)$

Step-by-step explanation:

Let's find the roots of every function:

$x(x-6)=0$

Expand the left side:

$x^2-6x=0$

Using quadratic formula:

$ax^2+bx+c=0\\x=\frac{-b\pm \sqrt{b^2-4ac} }{2a}$

Therefore:

$x=\frac{-(-6)\pm \sqrt{36-(4*1*0)} }{2*1} =\frac{6\pm \sqrt{36} }{2}=\frac{6\pm 6 }{2}$

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.