Which of the following describes the graph of h(x) = –2(–x + 3) + 4?

A. Exponential function decreasing from the left asymptotic to the line y equals 4 and going through the point 0 comma negative 4 down to the right
B. Exponential function increasing from the left and going through the point 0 comma negative 4 up to the right asymptotic to the line y equals 4
C. Exponential function decreasing from the left asymptotic to the line y equals negative 4 and going through the point negative 1 comma negative 8 down to the right
D. Exponential function increasing from the left and going through the point 1 comma negative 8 up to the right asymptotic to the line y equals negative 4

Inside the function notation, we have [tex]-x+3[/tex]. My recommendation is to factor that into [tex]-(x-3)[/tex], so I can read the transformations left-to-right.

This gives us [tex]h(x) = -f\big(-(x-3)\big)+4[/tex].

Things we know about the parent function [tex]f(x) =2^x[/tex]:

It increases from left-to-right.
It has a horizontal asymptote at y=0.
It goes through the point (0,1) and (1,2).

The transformations [tex]h[/tex] has done to [tex]f[/tex], we'd have:

Reflect over the y-axis, from the negative in front of [tex]f[/tex].
Reflect over the x-axis, from the negative inside [tex]f( ~~ )[/tex]
Shift right 3 units, from the "-3" inside.
Shift up 4 units, from the "+4" outside.

The two reflections mean you'll still have an increasing function, but it'll be under the x-axis instead of above the x-axis. It'll go from low in Quadrant III to close to the x-axis in Quadrant IV.

The means it's either B or D.

Since the entire graph is shifted up by 4 units, the horizontal asymptote will be moved from the x-axis (AKA y=0) to y=4.

The narrows it does to B.

To confirm, evaluate h(0) to make sure the graph goes through (0,-4).

[tex]\begin{aligned}h(0) &= -2^{(-0+3)}+4\\&= -2^{3}+4\\&=-8+4\\&= -4\end{aligned}[/tex]

That confirms it. Answer is B.