I hope my solutions are helpful.
7) Transformed equation has the form y-k=a(x-h)4, so we have to find a, h and k.
The turning-point of y=x4 occurs at (0,0) but the transformation shifts this (-2,5), therefore h=-2 and k=5.
y-5=a(x+2)4, y=a(x+2)4+5.
f(0)=58 (y-intercept), 58=16a+5, a=53/16, so y=53(x+2)4/16+5.
9) f(x)=-3(2x-10)4+243.
a) The x-intercepts are found by solving f(x)=0.
(2x-10)4=81, 2x-10=3 or -3 for real solutions. 2x=13, x=13/2; or 2x=7, x=7/2.
b) Range is (-∞,243] (that is, y≤243), because the maximum is when 2x-10=0 (x=5), all other values being less than 243.
c) Turning point at (5,243)--maximum (see (b))
d) Intervals of decrease for x [5,∞), that is, x≥5.
10) 5 is the compression factor. But by taking 5⅓ (or ∛5) inside the parentheses as a common factor, g(x) remains the same but the compression factor is effectively removed.
g(x)=(-2x∛5/3+24∛5)3-6. This is g(x) simplified. If we just remove the 5 from g(x) the graph is "wider" but has the same point of inflection, so it's not a simplification of g(x), but rather a different graph. There are other interpretations of this question. g(x)=5(⅔(-x+36))3-6=(40/27)(-x+36)3-6. 40/27 is the compression factor. If it is removed, it changes g(x) to (-x+36)3-6. Again, this not a simplification of g(x), it's a different function which resembles g(x).
11) -7y+3=√(2x-5), y=(3-√(2x-5))/7.
12) Let y=f(x)=½x-8, then x=f-1(y)=2y+16, f-1(x)=2x+16
h(f-1(x))=-3g(½f-1(x)-8)+4=-3g(x)+4.
So g(x)=⅓(4-h(f-1(x)))=⅓(4-h(2x+16)).
a) To map a point (x,y) from g to h, the mapping function x→2x+16 applies, y→-3g+4.
b) Since x>-5, 2x+16=6, making the domain of h {x∈ℝ| x>6}
c) -14<g<27: g>-14, -g<14, -3g<42, -3g+4<46; g<27, -g>-27, -3g>-81, -3g+4>-77; h(x)>-77, h(x)<46, so the range is {h(x)∈ℝ| -77<h(x)<46}