Question

the table shows values of a square root function, f(x).

x	0	1	4	9	16
f(x)	0	1	2	3	4

the equation shown represents a quadratic function, g(x).
$g(x) = (x - 2)^2 - 4$
compare the y-intercepts and minimum values of functions f(x) and g(x).
move options to the blanks to complete the sentences.
the y-intercept of f(x) is ______ the y-intercept of g(x).
the minimum value of f(x) is ______ the minimum value of g(x).
options: greater than, less than, equal to

Explanation:

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

The y-intercept occurs at \( x = 0 \). From the table, when \( x = 0 \), \( f(0) = 0 \).

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

For \( g(x)=(x - 2)^2-4 \), set \( x = 0 \):
\( g(0)=(0 - 2)^2-4 = 4 - 4 = 0 \).
So, y-intercepts of \( f(x) \) and \( g(x) \) are equal.

Step3: Find minimum value of \( f(x) \)

\( f(x) \) is a square - root function (\( f(x)=\sqrt{x} \) from the table: \( \sqrt{0}=0,\sqrt{1}=1,\sqrt{4}=2,\sqrt{9}=3,\sqrt{16}=4 \)). The square - root function \( y = \sqrt{x} \) has a domain \( x\geq0 \) and range \( y\geq0 \). So the minimum value of \( f(x) \) is \( 0 \).

Step4: Find minimum value of \( g(x) \)

\( g(x)=(x - 2)^2-4 \) is a quadratic function in vertex form \( y=a(x - h)^2 + k \), where \( a = 1>0 \), so the parabola opens upward and the vertex \( (h,k)=(2,-4) \) is the minimum point. The minimum value of \( g(x) \) is \( - 4 \).

Since \( 0>-4 \), the minimum value of \( f(x) \) is greater than the minimum value of \( g(x) \).

Answer:

The y - intercept of \( f(x) \) is equal to the y - intercept of \( g(x) \).
The minimum value of \( f(x) \) is greater than the minimum value of \( g(x) \).

So the first blank is "equal to" and the second blank is "greater than".