Question

6 the quadratic function ( f ) is represented by the equation, and the table gives some values of the quadratic function ( g ).
( f(x) = x^2 - 6x + 16 )

( x )	0	2	4	6	8

which statements are true?
select all that apply.
a the ( y )-intercept of ( g(x) ) is greater than the ( y )-intercept of ( f(x) ).
b the ( y )-coordinate of the vertex of ( g(x) ) is greater than the ( y )-coordinate of the vertex of ( f(x) ).
c the ( x )-coordinate of the vertex of ( g(x) ) is greater than the ( x )-coordinate of the vertex of ( f(x) ).
d the ( x )-coordinate of the vertex of ( f(x) ) is greater than the ( x )-coordinate of the vertex of ( g(x) ).
e the ( y )-intercept of ( f(x) ) is greater than the ( y )-intercept of ( g(x) ).
f the ( y )-coordinate of the vertex of ( f(x) ) is greater than the ( y )-coordinate of the vertex of ( g(x) ).

Explanation:

Step1: Find y-intercept of $f(x)$

For $f(x)=x^2+6x+16$, set $x=0$:
$f(0)=0^2+6(0)+16=16$

Step2: Find y-intercept of $g(x)$

From table, $x=0$ gives $g(0)=23$

Step3: Compare y-intercepts

$23>16$, so A is true, E is false.

Step4: Find vertex of $f(x)$

For $ax^2+bx+c$, vertex $x=-\frac{b}{2a}$:
$x=-\frac{6}{2(1)}=-3$
Substitute $x=-3$ into $f(x)$:
$f(-3)=(-3)^2+6(-3)+16=9-18+16=7$
Vertex of $f(x)$: $(-3, 7)$

Step5: Find vertex of $g(x)$

Table is symmetric about $x=4$ (since $g(2)=g(6), g(0)=g(8)$). Minimum $g(x)=5$ at $x=4$.
Vertex of $g(x)$: $(4, 5)$

Step6: Compare x-coordinates of vertices

$4>-3$, so C is true, D is false.

Step7: Compare y-coordinates of vertices

$7>5$, so F is true, B is false.

Answer:

A. The y-intercept of $g(x)$ is greater than the y-intercept of $f(x)$.
C. The x-coordinate of the vertex of $g(x)$ is greater than the x-coordinate of the vertex of $f(x)$.
F. The y-coordinate of the vertex of $f(x)$ is greater than the y-coordinate of the vertex of $g(x)$.