Q:

Given the functions f(x) = 4x2 − 1, g(x) = x2 − 8x + 5, and h(x) = –3x2 − 12x + 1, rank them from least to greatest based on their axis of symmetry.

Accepted Solution

A:
the axis of symmetry is x=-b/2a
for f(x), the axis of symmetry is x=0, because there is no b.
when x=0, f(x)=4*0²-1=-1, so the vertex is (0, -1)
because a in this case is 4, a positive number, the parabola opens upward, -1 is the lowest value.

for g(x), x=-(-8)/2*1=4, f(x)=4²-8*4+5=-11, the vertex is (4,-11)
because a is 1 in this case, a positive number, the parabola opens upward, -11 is the lowest value.

for h(x), x=-(-12)/[2*(-3)]=-2, f(x)=-3*(-2)²-12(-2)+1=13, the vertex is (-2,13)
because a is -3 in this case, a negative number, 13 is the largest value. 

f(x)'s is obviously larger than g(x) (the graph of f(x) is above that of g(x)), but for h(x), since it opens downward from y=13, it overwraps with part of f(x) and g(x), I'm not sure how you can compare that.

but if we look at the vertex alone, g(x) is the least, then f(x), then h(x) is the largest. 
I hope all this makes sense.