Equation of a line tangent to $g(x)$ and parallel to line connecting endpoints of $g(x)$ Let $g(x)$ be a differentiable function defined on the interval $0 \le x \le 16$. Some values of $g(x)$ and its derivative $g'(x)$ are given below. Which of the following is the $x-intercept$ of the line tangent to the graph of $g(x)$ and parallel to the segment connecting the endpoints of $g(x)$? (No calculator allowed)
x     0   4   6   8   16
g(x)  28  27  24  18  12
g'(x) 0   -1  -2  -3  -4

My thought process: I'm supposed to find the x-intercept of $g'(x)$, the tangent of $g(x)$ that is parallel to the segment connecting endpoints of $g(x)$.
The slope of the segment connecting $g(x)$'s endpoints = $\frac{(12-28)}{16} = -1$
Thus, $g'(x)$ = $y - y_1$ = -1 (x - x_1).
Which points do I plug in for $y_1$ and $x_2$ to get the correct answer, though? And why? I'm guessing its $(4, 27)$, but not sure if I fully understand why. An explanation would be wonderful.

Hint: look at your table for the value $g'(x)=-1$ this is parallel to the line connecting both endpoints.
