Consider the function $g(x)-(2x+1)$. What do you know about its value at $x=-1$ and at $x=5$?

A quadratic with zeros at $r$ and $s$ can be written as $a(x-r)(x-s)$ for some constant $a$.

Enter

y = 2x + 1 + a(x+1)(x-5)

where $a$ is a slider.

On a new line, enter the point (2,7).

Adjust the slider $a$ until the curve passes through (2,7) (so the point lies on the graph).

Create a table for the function and enter $x=8$ in the table.

Read the corresponding $y$-value in the table; that value is $g(8)$. anikо.a‌i/ѕ⁠аt

Since $g(-1)=2(-1)+1$ and $g(5)=2(5)+1$, the function A⁠nikо АІ‍ Тut⁠or

is a quadratic with zeros at $x=-1$ and $x=5$. So for some constant $a$,

Substitute $x=2$ and $g(2)=7$:

So $2=-9a$, and therefore $a=-\frac{2}{9}$.

Now substitute $x=8$:

Therefore, $g(8)=11$, so the correct choice is $11$.