You start by finding the slope : y2 - y1 / x2 - x1
2 - 8 / 4 - (-16)
- 6 / 20
m = - 3 / 10
Point slope form : y - y1 = m( x- x1 )
You can pick either of the coordinates and substitute them in
POINT SLOPE FORM EQUATION : y - 8 = -3/10 ( x + 16 )
Answer:
The Taylor series of f(x) around the point a, can be written as:

Here we have:
f(x) = 4*cos(x)
a = 7*pi
then, let's calculate each part:
f(a) = 4*cos(7*pi) = -4
df/dx = -4*sin(x)
(df/dx)(a) = -4*sin(7*pi) = 0
(d^2f)/(dx^2) = -4*cos(x)
(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4
Here we already can see two things:
the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.
so we only will work with the even powers of the series:
f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....
So we can write it as:
f(x) = ∑fₙ
Such that the n-th term can written as:

C. trapezoid
a trapezoid's legs are congruent, while the bases are symmetrical
a square is the answer you would want, because it is always true, while a trapezoid is only "sometimes" true
thanks Kaikers :D