Question

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. A) g(x), h(x), f(x)
B) f(x), h(x), g(x)
C) g(x), f(x), h(x)
D) h(x), f(x), g(x)

Answer 1

Answer:

Option D) h(x), f(x), g(x)

Step-by-step explanation:

we know that

The axis of symmetry of a vertical parabola is equal to the x-coordinate of the vertex of the parabola

Part 1) we have

$f(x)=4x^{2} -1$

This is a vertical parabola open upward

The vertex is a minimum The vertex is the point (0,-1)

The x-coordinate of the vertex is 0

so

The axis of symmetry is x=0

Part 2) we have

$g(x)=x^{2}-8x+5$

This is a vertical parabola open upward

The vertex is a minimum

Convert the equation into vertex form

$g(x)-5=x^{2}-8x$

$g(x)-5+16=x^{2}-8x+16$

$g(x)+11=x^{2}-8x+16$

$g(x)+11=(x-4)^{2}$

$g(x)=(x-4)^{2}-11$

The vertex is the point (4,-11)

The x-coordinate of the vertex is 4

so

The axis of symmetry is x=4

Part 3) we have

$h(x)=-3x^{2}-12x+1$

This is a vertical parabola open downward

The vertex is a maximum

Convert the equation into vertex form

$h(x)-1=-3x^{2}-12x$

$h(x)-1=-3(x^{2}+4x)$

$h(x)-1-12=-3(x^{2}+4x+4)$

$h(x)-13=-3(x+2)^{2}$

$h(x)=-3(x+2)^{2}+13$

The vertex is the point (-2,13)

The x-coordinate of the vertex is -2

so

The axis of symmetry is x=-2

Part 4) Rank their axis of symmetry from least to greatest

1) h(x) -----> axis of symmetry -2

2) f(x) -----> axis of symmetry 0

3) g(x) -----> axis of symmetry 4

so

h(x),f(x),g(x)