Question

a function g(x) has x-intercepts at (\frac{1}{2}, 0) and (6, 0). which could be g(x)?\
\bigcirc\\ g(x) = 2(x + 1)(x + 6)\
\bigcirc\\ g(x) = (x - 6)(2x - 1)\
\bigcirc\\ g(x) = 2(x - 2)(x - 6)\
\bigcirc\\ g(x) = (x + 6)(x + 2)

Explanation:

Step1: Recall x - intercept form

If a function \( g(x) \) has x - intercepts at \( x = a \) and \( x = b \), then the factored form of the function is \( g(x)=k(x - a)(x - b) \), where \( k\) is a non - zero constant.
The x - intercepts of \( g(x) \) are at \( (\frac{1}{2},0) \) and \( (6,0) \), so \( a=\frac{1}{2} \) and \( b = 6 \).

Step2: Analyze the factor for \( x=\frac{1}{2} \)

For the root \( x=\frac{1}{2} \), we can rewrite it as \( 2x-1 = 0\) (by solving \( x=\frac{1}{2}\) for the factor: \( 2x-1=0\Rightarrow x = \frac{1}{2}\)). For the root \( x = 6\), the factor is \( (x - 6) \) (since \( x-6=0\Rightarrow x = 6\)).
So the function in factored form should be \( g(x)=k(x - 6)(2x - 1) \), where \( k\) is a non - zero constant. When \( k = 1\), the function is \( g(x)=(x - 6)(2x - 1) \).
Let's check the other options:

• For \( g(x)=2(x + 1)(x + 6) \), the roots are \( x=-1\) and \( x=-6\), which are not the given roots.
• For \( g(x)=2(x - 2)(x - 6) \), the roots are \( x = 2\) and \( x=6\), \( x = 2\) is not the given root \( x=\frac{1}{2}\).
• For \( g(x)=(x + 6)(x + 2) \), the roots are \( x=-6\) and \( x=-2\), which are not the given roots.

Answer:

\( \boldsymbol{g(x)=(x - 6)(2x - 1)} \) (the second option)