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)

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nobillionaireNobley nobillionaireNobley · Answer 1 · 2016-06-30T19:03:19-04:00

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

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carlos2112 carlos2112 · Answer 2 · 2019-10-31T22:11:12-04:00

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.